According to this question, the exponential of an operator is defined as
$$ e^A = I + A + \frac{1}{2}A^2 + \frac{1}{3!}A^3 + \frac{1}{4!}a^4 + ...$$
Consider the following equation:
$$ i \frac{\partial}{\partial t}e^{iAt}\Psi = Be^{iAt}\Psi$$
Then comparing both sides,
$$ LHS = i \frac{d}{dt}\left(\Psi + iAt\Psi + \frac{1}{2}(iAt)^2\Psi + \frac{1}{3!}(iAt)^3\Psi + \frac{1}{4!}(iAt)^4\Psi + ...\right)$$
$$ = i \frac{d}{dt}(\Psi) - A\frac{d}{dt}(t\Psi) - i \frac{A^2}{2} \frac{d}{dt}(t^2\Psi) + \frac{A^3}{3!}\frac{d}{dt}(t^3\Psi) + i \frac{A^4}{4!}\frac{d}{dt}(t^4\Psi) + ...$$
$$ = i \frac{d\Psi}{dt} - A(t\frac{d\Psi}{dt} + \Psi) - i \frac{A^2}{2} (t^2\frac{d\Psi}{dt} + 2t\Psi) + \frac{A^3}{3!}(t^3\frac{d\Psi}{dt} + 3t^2\Psi) + i \frac{A^4}{4!}(t^4\frac{d\Psi}{dt} + 4t^3\Psi) + ...$$
$$ = (-A - iA^2t + \frac{A^3t^2}{2!} + \frac{iA^4t^3}{3!} + ...)\Psi + (i - At - \frac{iA^2t^2}{2} + \frac{A^3t^3}{3!} + \frac{iA^4t^4}{4!} + ...)\frac{d\Psi}{dt}$$
and
$$ RHS = B\left(I + iAt + \frac{1}{2}(iAt)^2 + \frac{1}{3!}(iAt)^3 + \frac{1}{4!}(iAt)^4 + ...\right)\Psi$$
$$ = (B + iBAt - \frac{1}{2}BA^2t^2 - i\frac{1}{3!}BA^3t^3 + \frac{1}{4!}BA^4t^4 + ...)\Psi$$
I am unsure what is happening with these infinite sums on both sides, and what is the purpose/utility of writing an exponential operator equation in this way?
In the present case, you needn't use such expansions, given that the solution is deduced straightforwardly for $\Phi := e^{iAt}\Psi$ as $\Phi(t) = e^{-iBt}\Phi_0$, with $\Phi_0 = \Phi(0)$, hence $\Psi(t) = e^{-iAt}\Phi(t) = e^{-iAt}e^{-iBt}e^{iAt}\Psi_0$. The Taylor expansion only serves as a definition for the exponential of an operator/matrix if you would like to compute it explicitly.