In what sense is the proper/effective action $Gamma[phi_c]$ a quantum-corrected classical action $S[phi]$?

We want to calculate the path integral $$ Z = int mathcal{D}{phi}, e^{i hbar^{-1} S[phi]} $$ which encodes a transition amplitude between initial and final quantum states.

If we had the effective action $Gamma[phi]$ at our disposal, we would have calculated the same result by solving for $$ phi_c(x):quad left. frac{delta Gamma}{delta phi} right|_{phi=phi_c} = 0 $$ and plugging it back in the effective action: $$ Z = e^{i hbar^{-1} Gamma[phi_c]}. $$

This is the definition of $Gamma$.

Note that no path integrals are required at this point. Boundary conditions are implicitly present throughout this answer, encoding the exact states between which the quantum transition occurs. Their existence ensures that there is only one solution $phi_c$.

Now to why $phi_c$ is called classical: it solves the e.o.m. given by the action $Gamma$.

Think of $Gamma$ as of an object in which all the short-scale properties of the integration measure $mathcal{D}phi$ (including renormalization-related issues) are already accounted for. You simply solve the e.o.m. and plug the solution in the exponential and you are done: here is your transition amplitude.

That being said, $Gamma$ is not classical in the sense that it still describes dynamics of a quantum theory. Only in a different fashion. Simple algebraic manipulations instead of path integrals.

Finally, note how if the path integral is Gaussian, $$Gamma[phi] = S[phi] + text{const},$$ where $text{const}$ accounts for the path integral normalization constant. There are no quantum corrections.

In classical theory, however, we solve the e.o.m. w.r.t. $phi = phi_c$ for $S[phi]$, not $Gamma[phi]$. Plugging it back into $S[phi_c]$ gives us the Hamilton function. When the path integral is Gaussian, it doesn't matter if we use $S$ or $Gamma$, and exponentiating the Hamilton function gives you the transition amplitude. However if we are dealing with an interacting theory, the correct way to do this would be to use $Gamma$ instead of $S$. In this sense, $Gamma$ is the quantum-corrected version of $S$.

And yes, it is always true (can be shown using the saddle point approximation formula) that $$ Gamma[phi] = S[phi] + mathcal{O}(hbar). $$

Why wouldn't we just use $Gamma[phi]$ to define the quantum theory and forget about $S[phi]$ alltogether? Because $Gamma$ is non-local and contains infinitely many adjustable parameters. These can be determined from the form of $S[phi]$ by, well, quantization. That's why it is $S[phi]$ which defines the theory, not $Gamma$. $Gamma$ is to be calculated via path integrals.

UPDATE: It is also important to understand that in naive QFT $Gamma$ contains divergences, while $S$ doesn't. However, the actual situation is opposite. It is $S$ which contains divergences (divergent bare couplings), which cancel out against the divergences coming from the path integral, rendering a finite (i.e. renormalized) $Gamma$. That $Gamma$ should be finite is evident from how we use it to calculate physical properties: we only solve the e.o.m. and plug the result back in $Gamma$.

Actually, the whole point of renormalization is to make $Gamma$ finite and well-defined while adjusting only a finite number of diverging couplings in the bare action $S$.

There is already a good answer by Solenodon Paradoxus. Here we provide a formal proof (via the stationary phase/WKB approximation).

1. To fix notation, we define the effective/proper action $$ Gamma[phi_{rm cl}]~=~W_c[J]-J_k phi_{rm cl}^k, tag{1}$$ as the Legendre transformation of the generating functional $W_c[J]$ for connected diagrams. We assume that the Legendre transformation is regular, i.e. the formula $$ phi_{rm cl}^k~=~frac{delta W_c[J]}{delta J_k} qquad Leftrightarrow qquad J_k~=~-frac{delta Gamma[phi_{rm cl}]}{delta phi_{rm cl}^k} tag{2}$$ is invertible. Here $J_k$ are the sources and $phi_{rm cl}^k$ are the so-called classical fields. (The latter terminology is a bit of a misnormer as $phi_{rm cl}^k[J]$ as a function of the sources $J_{ell}$ could depend explicitly on $hbar$. See also section 8 below.)

2. The partition function/path integral is $$ expleft{ frac{i}{hbar} W_c[J]right}~=~Z[J] ~:=~int ! {cal D}frac{phi}{sqrt{hbar}}~expleft{ frac{i}{hbar} left(S[phi]+J_k phi^kright)right} . tag{3}$$ The first equality in eq. (3) is the linked cluster theorem, cf. e.g. this Phys.SE post.

3. At this place it is customary to mention some elementary facts. The 1-pt function/quantum averaged field is by definition $$begin{align} langle phi^k rangle_J &~:=~frac{1}{Z[J]} int ! {cal D}frac{phi}{sqrt{hbar}}~phi^kexpleft{ frac{i}{hbar} left(S[phi]+J_{ell} phi^{ell}right)right}cr &~=~frac{1}{Z[J]} frac{hbar}{i} frac{delta }{delta J_k}int ! {cal D}frac{phi}{sqrt{hbar}}~expleft{ frac{i}{hbar} left(S[phi]+J_{ell} phi^{ell}right)right}cr &~stackrel{(3)}{=}~frac{1}{Z[J]} frac{hbar}{i}frac{delta Z[J]}{delta J_k}~stackrel{(3)}{=}~frac{delta W_c[J]}{delta J_k}~stackrel{(2)}{=}~phi_{rm cl}^k, end{align} tag{4}$$

4. The 2-pt function is by definition $$begin{align} langle phi^k phi^{ell}rangle_J &~:=~frac{1}{Z[J]} int ! {cal D}frac{phi}{sqrt{hbar}}~phi^kphi^{ell}expleft{ frac{i}{hbar} left(S[phi]+J_m phi^mright)right}cr &~stackrel{(3)}{=}~frac{1}{Z[J]} left(frac{hbar}{i}right)^2frac{delta^2 Z[J]}{delta J_kdelta J_{ell}}~stackrel{(3)}{=}~frac{1}{Z[J]} frac{hbar}{i} frac{delta}{delta J_k} left(Z[J]frac{delta W_c[J]}{delta J_{ell}}right)cr &~stackrel{(4)}{=}~frac{hbar}{i} frac{delta^2 W_c[J]}{delta J_kdelta J_{ell}} + langle phi^k rangle_J langle phi^{ell} rangle_J,end{align} tag{5}$$ i.e. the connected 2-pt function plus a disconnected piece.

5. Now let us return to OP's question. By formal Fourier transformation of the path integral (3), we get $$ expleft{ frac{i}{hbar}S[phi_{rm cl}]right} ~stackrel{(3)}{=}~int ! {cal D}frac{J}{sqrt{hbar}}~expleft{ frac{i}{hbar} left(W_c[J]-J_k phi_{rm cl}^kright)right} $$ $$~stackrel{(1)}{sim}~ {rm Det}left(frac{1}{i}frac{delta^2 W_c[J[phi_{rm cl}]]}{delta J_k delta J_{ell}}right)^{-1/2} expleft{ frac{i}{hbar}Gamma[phi_{rm cl}]right}left(1+ {cal O}(hbar)right) $$ $$~stackrel{(8)}{=}~ {rm Det}left(frac{1}{i}frac{delta^2 Gamma[phi_{rm cl}]}{delta phi_{rm cl}^k delta phi_{rm cl}^{ell}}right)^{1/2} expleft{ frac{i}{hbar}Gamma[phi_{rm cl}]right}left(1+ {cal O}(hbar)right) quadtext{for}quadhbar~to~0 tag{6}$$ in the stationary phase/WKB approximation. In the last equality of eq. (6), we used that $$begin{align}delta^k_{ell} ~=~frac{delta phi_{rm cl}^k[J[phi_{rm cl}]]}{deltaphi_{rm cl}^{ell}} ~=~&frac{delta phi_{rm cl}^k[J[phi_{rm cl}]]}{delta J_m} frac{delta J^m[phi_{rm cl}]}{deltaphi_{rm cl}^{ell}} cr ~stackrel{(2)}{=}~& -frac{delta^2 W_c[J[phi_{rm cl}]]}{delta J_kdelta J_m} frac{delta^2 Gamma[phi_{rm cl}]}{deltaphi_{rm cl}^mdeltaphi_{rm cl}^{ell}},end{align} tag{7}$$ i.e.

   $$text{The 2-pt functions } frac{1}{i}frac{delta^2 W_c[J]}{delta J_kdelta J_m} text{ and } frac{1}{i}frac{delta^2 Gamma[phi_{rm cl}]}{deltaphi_{rm cl}^mdeltaphi_{rm cl}^{ell}} text{ are inverses of each other.} tag{8}$$

6. Eq. (6) shows that the effective action $$begin{align} Gamma[phi_{rm cl}] ~stackrel{(6)}{=}~& S[phi_{rm cl}] +frac{ihbar}{2}ln {rm Det}left(frac{1}{i}frac{delta^2 Gamma[phi_{rm cl}]}{delta phi_{rm cl}^k delta phi_{rm cl}^{ell}}right) +{cal O}(hbar^2) tag{9} cr ~stackrel{(9)}{=}~& S[phi_{rm cl}] +frac{ihbar}{2}ln {rm Det}left(frac{1}{i} H_{kell}[phi_{rm cl}]right) +{cal O}(hbar^2) tag{10}end{align}$$ agrees with the action $S$ up to quantum corrections. In eq. (10) we have defined the Hessian $$ H_{kell}[phi]~:=~ frac{delta^2 S[phi]}{deltaphi^kdeltaphi^{ell}}. tag{11} $$ (The square root factor in eq. (6) only contributes at one-loop and beyond.)

   In other words, if we assume that the action $S$ has no explicit $hbar$-dependence, we deduce that to zeroth-order in $hbar$/tree diagrams in the effective action

   $$ Gamma_{text{tree}}[phi_{rm cl}] ~stackrel{(9)}{=}~S[phi_{rm cl}] tag{12}$$

   is equal to the action $S$ itself. Similarly, we deduce that to first-order in $hbar$/one-loop diagrams in the effective action

   $$ Gamma_{text{1-loop}}[phi_{rm cl}] ~stackrel{(10)}{=}~frac{ihbar}{2}ln {rm Det}left(frac{1}{i} H_{kell}[phi_{rm cl}] right) tag{13}$$

   is equal to a functional determinant of the Hessian of the action $S$. Eqs. (10), (12) & (13) answer OP's question. See also this related Phys.SE post.

7. At this place it is customary to mention some elementary facts. Let there be given fixed sources $J_k$. From$^1$ $$frac{delta Gamma[phi_{rm cl}]}{delta phi_{rm cl}^k}~stackrel{(2)}{=}~-J_k~stackrel{text{EL eqs.}}{approx}~frac{delta S[phi_0]}{delta phi^k}, tag{14} $$ we deduce that the so-called classical solution $phi_{rm cl}^k$ and the Euler-Lagrange (EL) solution $phi_0^k$ agree$^1$ $$ phi_{rm cl}^k[J]~stackrel{(9)+(14)}{approx}~phi_0^k[J] +{cal O}(hbar) tag{15} $$ up to quantum corrections. Eq. (15) justifies the practice to call $phi_{rm cl}^k$ the classical field. (We assume that each solution to eq. (14) is unique, due to pertinent boundary conditions. We have excluded instantons for simplicity.)

8. Alternatively, from the background field method $$ underbrace{phi^k}_{text{quan. field}} ~=~overbrace{underbrace{phi^k_{rm cl}}_{text{clas. field}}}^{text{backgr. field}}+underbrace{eta^k}_{text{fluctuation}}, tag{16}$$ the effective action (1) becomes $$expleft{frac{i}{hbar}Gamma[phi_{rm cl}]right} ~stackrel{(1)+(3)}{=}~ int!{cal D}frac{phi}{sqrt{hbar}} ~expleft{frac{i}{hbar} left(S[phi] +J_k[phi_{rm cl}](phi^k-phi^k_{rm cl}) right) right} $$ $$~stackrel{(16)}{=}~ int!{cal D}frac{eta}{sqrt{hbar}} ~expleft{frac{i}{hbar} left(S[phi_{rm cl}+eta] +J_k[phi_{rm cl}] eta^k right)right} $$ $$~=~ int!{cal D}frac{eta}{sqrt{hbar}} ~expleft{frac{i}{hbar} left( S[phi_{rm cl}] +underbrace{left(frac{delta S[phi_{rm cl}]}{delta phi_{rm cl}^k} +J_k[phi_{rm cl}]right)}_{={cal O}(hbar)} eta^k +frac{1}{2}eta^k H_{kell}[phi_{rm cl}] eta^{ell} +{cal O}(eta^3) right)right} $$ $$~stackrel{text{WKB approx.}}{sim}~ {rm Det}left(frac{1}{i}H_{mn}[phi_{rm cl}] right)^{-1/2}left(1+ {cal O}(hbar)right) $$ $$ times expleft{ frac{i}{hbar}left(S[phi_{rm cl}] -frac{1}{2}underbrace{left(frac{delta S[phi_{rm cl}]}{delta phi_{rm cl}^k} +J_k[phi_{rm cl}]right)}_{={cal O}(hbar)} (H^{-1})^{kell}[phi_{rm cl}] underbrace{left(frac{delta S[phi_{rm cl}]}{delta phi_{rm cl}^{ell}} +J_{ell}[phi_{rm cl}] right)}_{={cal O}(hbar)} right)right} $$ $$~stackrel{(2)+(15)}{=}~ {rm Det}left(frac{1}{i}H_{mn}[phi_{rm cl}]right)^{-1/2}expleft{ frac{i}{hbar}S[phi_{rm cl}]right}left(1+ {cal O}(hbar)right) quadtext{for}quadhbar~to~0 tag{17}$$ in the stationary phase/WKB approximation. Eq. (17) again leads to the sought-for eq. (10).

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$^1$ The $approx$ symbol means here equality modulo the Euler-Lagrange (EL) equations.

Obviously, $phi(x)$ is different from $phi_c(x)$. The former is a classical field of a classical field theory, the latter is just a quantity that appears in the Legendre transform of the generating functional for connected Green functions. It just happens that for classical actions that an be treated as perturbations around quadratic actions, the equations satisfied by $phi_c(x)$ coincide with those of $phi(x)$ in classical field theory, at the limit $hbarrightarrow 0$.

Except for the suggestive name, there is also no quantum-classical correspondence: $phi_c(x)$ is not the expectation value of the field $hat{phi}(x)$ in presence of an external source (expressed in terms of properly defined probabilities). It does not make sense as a quasi-classical observable.

Furthermore, the effective action is non-local, and hence, it does not generate any effective quasi-classical dynamics. The effective action is only a generator for Green functions relevant to the calculation of S-matrix elements.