The thing you wrote with the operator $C = D \otimes E$ is correct.
The thing you may have missed is that the operator $H_0$ is the sum of two terms $H_{1,2}$, each of which acts in different subspaces of the full Hilbert space. This means that the two terms $H_{1,2}$ commute with each other. So when we write $\exp (\lambda (A + B))$ for $[A,B]=0$, we can just write this as $ e^{\lambda (A+B)} = e^{\lambda A} \, e^{\lambda B} = e^{\lambda B}\, e^{\lambda A}$. If $[A,B]\neq 0$, then we use the Baker-Campbell-Hausdorff formula.
In other words, $\exp (- i H_0 t) = \exp (- i H_1 t) \, \exp (-i H_2 t)$. This might be all you need to move forward, but just in case, here's the rest:
This whole thing is even easier if you define the Pauli operator $Z = \left| g \middle\rangle \hspace{-0.3mm} \middle\langle g \right| - \left| e \middle\rangle \hspace{-0.3mm} \middle\langle e \right|$ so that $H_1 = \hbar \omega_a Z / 2$ and $\exp (- i H_1 t / \hbar) = \cos (\omega_a t / 2) \, \mathbb{1} - i \, \sin (\omega_a t/2) \, Z$ and $\exp ( - i H_2 t / \hbar ) = \exp( - i \omega_p t \, N / 2 )$ up to a phase that will be cancelled by $\exp ( + i H_2 t / \hbar )$
Now, $\delta H = \alpha \, X \otimes (a + a^{\dagger})$, and we have:
$ e^{i \, H_0 \, t/\hbar} \, \delta H \, e^{-i \, H_0 \, t/\hbar} \, = \, \left[ \, e^{i \, \omega_a \, Z / 2} \, X \, e^{-i \, \omega_a \, Z / 2}\, \right] \otimes \left[ \, e^{i \,\omega_p \,t \, \hat{N}/2} \,\left( \hat{a} + \hat{a}^{\dagger} \right) \, e^{-i \,\omega_p \,t \, \hat{N}/2} \, \right] $
$= \left[\, \left( \cos^2 (\frac{\omega_a \, t}{2} )- \sin^2 (\frac{\omega_a \, t}{2} ) \right) \, X + i \, \sin (\frac{\omega_a \, t}{2} )\, \cos (\frac{\omega_a \, t}{2} ) \, \left[ Z, X \right] \, \right] \otimes \sum\limits_{n=0}^{\infty} \frac{(i \, \omega_p / 2)^n}{n!} \, \left[ \hat{N}, \hat{a} + \hat{a}^{\dagger} \right]_n $
where I used the Campbell identity (one of the BCH guys), where $[A,B]_0 = B$, $[A,B]_1 = [A,B]$, $[A,B]_2 = [A,[A,B]]$, and so on. We then use $[\hat{N},\hat{a}]=-\hat{a}$ and $[\hat{N},\hat{a}^{\dagger}]=\hat{a}^{\dagger}$ to find
$e^{i \, H_0 \, t/\hbar} \, \delta H \, e^{-i \, H_0 \, t/\hbar} \, = \, \left[\, \cos (\omega_a \, t) \, X - \sin (\omega_a \, t )Y \, \right] \otimes \left[ \, e^{- i \, \omega_p \, t/2} \, \hat{a} + e^{i \, \omega_p \, t/2} \, \hat{a}^{\dagger} \right]$
as the answer? Unless I made some algebraic mistakes.
In this case, it's easiest just to deal with the operators. Whenever you deal with tensor-product Hilbert spaces (or many-body quantum mechanics in general), it can be helpful to imagine implicit identities attached to every single- or few-body operator.
Best Answer
You actually already have the result: one can always define an operator $$\hat{\pi}_{ml}\equiv (\mathbb{I}_s\otimes \langle A|_a) \hat{U}^\dagger_{sa} (|m\rangle_s\otimes |l\rangle_a)(_a\langle l|\otimes \vphantom{a}_s\langle m|)\hat{U}_{sa}(|A\rangle_a\otimes \mathbb{I}_s),$$ where I just added some parentheses, added a subscript $s$ for the system Hilbert space, added the subscript $sa$ to the unitary because it acts on both, and added identity operations for the system. We simply match the parts for each subspace to obtain $$\hat{\pi}_{ml}=\mathbb{I}_s (\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a)^\dagger |m\rangle_s \vphantom{a}_s\langle m|(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a) \mathbb{I}_s=(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a)^\dagger |m\rangle_s \vphantom{a}_s\langle m|(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a) .$$ The operator in parentheses $(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a)$ is just an operator on the system.
As to the properties of $\hat{\pi}_{ml}$, one would have to prove more things, but that is not the focus of this question.
Why did we define it like this? We simply expanded the absolute square $|\langle a|B|c\rangle|^2=\langle a|B|c\rangle\langle c|B^\dagger|a\rangle=\langle c|B^\dagger|a\rangle\langle a|B|c\rangle$ to find $$\left|(_a\langle l|\otimes \vphantom{a}_s\langle m|)\hat{U}_{sa}(|A\rangle_a\otimes |\psi\rangle_s)\right|^2=(\langle\psi|_s\otimes \langle A|_a) \hat{U}^\dagger_{sa} (|m\rangle_s\otimes |l\rangle_a)(_a\langle l|\otimes \vphantom{a}_s\langle m|)\hat{U}_{sa}(|A\rangle_a\otimes |\psi\rangle_s).$$
Now, your main confusion probably comes from not being used to operators that act on two systems. So let's just be clear and say that any such operator can be decomposed as $$\hat{M}_{sa}=\sum_{ijkl}m_{ijkl}|i\rangle_s \vphantom{a}_s\langle j|\otimes |k\rangle_a\vphantom{a}_a\langle l|.$$ Making this operator unitary, or choosing whether it entangles two systems or not, depends on the coefficients $m_{ijkl}$, but such a decomposition always exists. This lets us compute things like, for $$\hat{U}_{sa}\equiv\sum_{ijkl}u_{ijkl}|i\rangle_s \vphantom{a}_s\langle j|\otimes |k\rangle_a\vphantom{a}_a\langle l|,$$ $$\vphantom{a}_a\langle l^\prime|\hat{U}_{sa}|A\rangle_a=\sum_{ijkl}u_{ijkl}|i\rangle_s \vphantom{a}_s\langle j| (\vphantom{a}_a\langle l^\prime|k\rangle_a\vphantom{a}_a\langle l|A\rangle_a).$$ This is just a new operator $$\hat{M}_s=\sum_{ij}m_{ij}|i\rangle_s\vphantom{a}_s\langle j|$$ acting on the system alone, with coefficients that depend on $l^\prime$ and $A$: $$m_{ij}=\sum_{kl}u_{ijkl} (\vphantom{a}_a\langle l^\prime|k\rangle_a\vphantom{a}_a\langle l|A\rangle_a).$$