How to see the ${rm SO}(4)$ symmetry of the classical Kepler problem?

The classical treatment is actually simpler than the quantum one, since there are no operator ordering problems. This is what Pauli relied on in his solution of the Hydrogen atom, before Schroedinger, as every decent physicist of his generation had worked out the Kepler problem like this.

The idea is to find as many independent constants of the motion Q as possible, and break them up into independent, ideally PB-commuting, quantities.

Every such invariant must PB-commute with the Hamiltonian, since that is its time derivative, $$ { Q,H }= 0.$$

As a result, the analog of your quantum (2) is $$ H'= e^{ theta {Q ,} H= H+theta { Q,H}+ {theta^2 over 2!} {Q,{Q,H }}+... =H. $$ The linear first order phase-space gradients ${Q, bullet }={ partial Qover partial x^i} {partial over partial p^i}- { partial Qoverpartial p^i } {partial over partial x^i} $ are the generators of Lie transformations in phase space. They close into a Lie algebra purely classically, via PB commutation, $$ {Q, {S, - {S, { Q ,= { {Q,S}, $$ by Jacobi's identity. They are thus Lie algebra elements, and can act adjointly.

For the specific Kepler problem, let me follow an older talk of mine.

$$ H = frac{{bf p}^2}{2}-frac{1}{r} ~, $$ in simplified (rescaled) notation.

The invariants of the hamiltonian are the angular momentum vector, $$ {bf L} = {bf r} times {bf p}~, $$ and the Hermann-Bernoulli-Laplace vector, now usually called the Pauli-Runge-Lenz vector, $$ {bf A}= {bf p} times {bf L} -hat{{bf r}}~. $$ (Dotting it by $hat{{bf r}}$ instantly yields Kepler's elliptical orbits by inspection, $hat{{bf r}}cdot {bf A}+1 = {bf L}^2 /r$.)

Since $~{bf A}cdot {bf L}=0$, it follows that $$ H=frac{{bf A}^2-1}{2{bf L}^2 } ~~. $$ However, to simplify the PB Lie-algebraic structure, $$ { L_i, L_j } = epsilon^{ijk} L_k, qquad { L_i, A_j } = epsilon^{ijk} A_k, qquad { A_i, A_j } =-2H epsilon^{ijk} L_k, $$ it is useful to redefine ${bf D} equiv frac{{bf A}}{sqrt{- 2 H}}$, and further $$ {cal R}equiv {bf L} + {bf D}, qquad {cal L}equiv {bf L} - {bf D}. $$ These six simplified invariants obey the standard $SU(2)times SU(2) sim SO(4)$ symmetry algebra, $$ { {cal R}_i, {cal R}_j } = epsilon^{ijk} {cal R}_k, qquad { {cal R}_i, {cal L}_j } = 0, qquad { {cal L}_i, {cal L}_j } =epsilon^{ijk} {cal L}_k, $$ and depend on each other and the hamiltonian through
$$ bbox[yellow]{H=frac{-1}{2 {cal R}^2}=frac{-1}{2 {cal L}^2} }~, $$ so only five of the invariants are actually algebraically independent.

It is then manifest that all six respective infinitesimal transformations leave their Casimirs invariant, the Casimirs of the other SU(2) invariant, and hence the Hamiltonian invariant; consequently the corresponding finite transformations do likewise, $$ H'= e^{ theta^i { {cal R}_i,} ~ e^{phi^j{ {cal L}_j ,} ~ H= H ~. $$

You could work out these explicit infinitesimal phase-space transformations involved, if you were really inclined, and Fock and Bargmann do that, as linked in the comments, but this is the whole point of systematic formalism, that it cuts through the confusion and forestalls missing the forest for the trees. It's Lie's major contribution to geometry.