= Solution
For the requested oscillator response, treat $y$ as the fixed transverse position of a one-dimensional <quantum harmonic oscillator> along $z$, and put $\theta=ky$. The supplied radiation is transverse and already satisfies the <Coulomb gauge>. Its perturbation is
$$
H_1(t)=\frac{eA_0}{m}p_z\left(e^{i\theta}e^{-i\omega t}+e^{-i\theta}e^{i\omega t}\right),\quad t>0.
$$
The <annihilation operator> and <creation operator> give
$$
p_z=i\sqrt{\frac{m\hbar\omega_0}{2}}(a^\dagger-a),\qquad
\langle n|p_z|0\rangle=i\sqrt{\frac{m\hbar\omega_0}{2}}\,\delta_{n1}.
$$
Thus the only first-order transition out of the <ground state> is to $|1\rangle$. More generally, the oscillator <electric-dipole selection rule> is $\Delta n=\pm1$. This is a rule for <matrix elements>; exact energy conservation is replaced by a finite-time resonance profile during a switched finite-duration interaction.
To display both the resonant and nonresonant terms without spurious singularities, define
$$
F_\nu(t)=\int_0^t e^{i\nu t'}dt'
=\begin{cases}(e^{i\nu t}-1)/(i\nu),&\nu\ne0,\\t,&\nu=0,\end{cases}
\qquad C= eA_0\sqrt{\frac{\omega_0}{2m\hbar}}.
$$
The <interaction picture> coefficient in the given expansion is then
$$
\boxed{b_1(t)=C\left[e^{i\theta}F_{\omega_0-\omega}(t)
+e^{-i\theta}F_{\omega_0+\omega}(t)\right],\qquad b_n(t)=0\ (n\geq2).}
$$
The full <Schrödinger picture> coefficient of $|1\rangle$ is $e^{-3i\omega_0t/2}b_1(t)$, while that of $|0\rangle$ is $e^{-i\omega_0t/2}$ at this order. The normalization changes only at order $A_0^2$. In particular, at $\omega=\omega_0$ the first term is $Ce^{i\theta}t$, finite for every finite time, and the perturbative result requires $|b_1|\ll1$.
The fixed-$y$ interpretation is needed for the displayed target formulae. If $p^2$ were instead a three-dimensional kinetic operator with no transverse confinement, that model would have no normalizable transverse <ground state>, and $e^{\pm iky}$ would transfer transverse <momentum> $\pm\hbar k$. A treatment of that different model needs specified transverse states and recoil energies; they are absent from the question. The following parts use the intended localized-oscillator response.
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