Expression of the time evolution operator $hat U(t)$ in terms of the hamiltonian

This process is valid, but indirectly so. Like, if you were my student and you handed it to me I would be very worried that you had a “magical” attitude towards the mathematics involved, whereas this is something more than just an “umbral” similarity between two different mathematical domains.

What is actually happening is that you have a time-independent Hamiltonian $hat H$ and therefore it has stationary eigenstates $hat H |nrangle = E_n |nrangle$. Under the Schrödinger equation these then gain a time dependence which multiplies them by a rotating phase factor, $$|n(t)rangle = e^{-i E_n t/hbar} |nrangle$$ so that for an arbitrary other state if $|Psi(0)rangle = sum_n c_n |nrangle$ then $$|Psi(t)rangle = sum_{n=0}^infty c_n~e^{-iE_n t/hbar} |nrangle.$$This operator is justifiably then $e^{-ihat H t/hbar};$ that is what you get when you do a Taylor series.

So, what you are doing here is you are performing this derivation in a basis in which the Hamiltonian is diagonal. That is the missing aspect of your argument which indirectly makes the whole thing work; it is that if you look at this equation $$i hbar frac{partial U}{partial t} = hat H U$$in the basis in which $hat H$ is diagonal then a diagonal $U$ suffices to solve it, as the product of two diagonal matrices is diagonal: and furthermore each term on the diagonal is a separate differential equation $i hbar ~dot U_n(t) = E_n ~U_n(t)$ with boundary condition $U_n(0) = 1.$ It is valid because it is obviously true in one particular basis.

It’s also worth pointing out that the order $hat H U$ does not matter here but it does matter when $hat H = hat H(t)$ is no longer constant over time, in which case you get a term which is often written, since we don't have a great notation for continuous products, as $$U(t) = mathcal T expleft[-frac{i}{hbar} int_0^tmathrm dtau~hat H(tau) right],$$ the symbol $mathcal T$ meant to remind us that this is to be interpreted as a time-ordered series of products $$U(t) = lim_{delta tto 0} e^{-ihat H(t-delta t)delta t/hbar}~e^{-ihat H(t-2delta t)delta t/hbar}dots e^{-ihat H(delta t)delta t/hbar}e^{-ihat H(0)delta t/hbar}.$$ Per the other question you asked, the adjoint of this operator does not just swap out $+i$ for $-i$ but also it must reverse the time ordering of these terms for an anti-time-ordering $bar{mathcal T}.$

In particular in an interaction picture we split the Hamiltonian into some “easy” part and some “interacting part” $hat H = hbar (eta_0(t) + xi(t))$ and then we try to invent the easy-evolution operator $u$ as, $$u(t)=mathcal T expleft[ -i int_0^t mathrm dtau~eta_0(tau)right].$$Now since $u u^dagger = 1$ by construction, we can insert it into all of our expectation values to get a sort of quantum coordinate transform, mapping $hat A to tilde A = u^dagger hat A u$ while mapping $|Psi(t)rangle to |tilde Psi(t)rangle = u^dagger |Psi(t)rangle$ obeying the new equations, $$i frac{mathrm dtilde A}{mathrm dt} = -[eta, tilde A] + frac{partial tilde A}{partial t}, \ i frac{mathrm dhphantom{t}}{mathrm dt} |tildePsi(t)rangle = tildexi~ |tildePsi(t)rangle. $$ Doing this derivation correctly absolutely requires that the order flip when calculating $mathrm d/mathrm dt(u^dagger),$ so that the $eta_0$ comes out on the right hand side of the operator. This requires keeping straight in your head that $u^dagger$ is anti-time-ordered and so $u^dagger(t + delta t)approx u^dagger(t) e^{i eta_0(t) delta t}$ appears on the right of the operator and can be expanded out to first-order on that side.