It is a perfectly well-defined expression because the tensor product is a linear space.
The vectors $|v\rangle\otimes |w\rangle$ form a basis of the whole tensor-product vector space, so any vector (including Bell's state) in this space may be written as linear combinations of such basis vectors.
$$ |\psi \rangle = \sum_{ij} c_{ij} |v_j\rangle\otimes |w_j\rangle $$
Because the operators are linear and we know how to act on each term, the result of the action of the operator $L$ is
$$ L |\psi \rangle = \sum_{ij} c_{ij} L |v_j\rangle\otimes |w_j\rangle $$
where your formulae already say how to evaluate the individual terms e.g. for $L = E\otimes I$.
The natural inner product of two vectors on the tensor product space is given by the simple product of the factors. Choose a basis like above, and write the inner product of two basis vectors as products in the most straightforward way.
$$ \langle v_i| \otimes \langle w_j| \cdot |v_m\rangle \otimes |w_k\rangle = \langle v_i|v_m\rangle \cdot \langle w_j|w_k\rangle$$
This again defines the inner product for any two vectors, by linearity. Decompose each of the two general vectors in the tensor product space that enter the inner product as a linear combination of the simple $vw$ basis vectors above, apply the distributive law to calculate the inner product of each term, and sum the terms with the same coefficients.
The two different Hamiltonian forms represent different physical things. The first, given by
$$
\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\tag{1},
$$
represents the free Hamiltonians of each system separately, and the second, given by
$$
\hat{H}_1\otimes\hat{H}_2\tag{2},
$$
represents an interaction term between two quantum systems. (So, e.g. the first might be the kinetic and potential energies of two individual oscillators, and then the second might represent the interaction Hamiltonian if, for instance, the two oscillators are connected by a spring).
Importantly, Hamiltonians of the form (1) evolve unentangled states into unentangled ones, but Hamiltonians of the form (2) can evolve unentangled states into entangled ones. To see this, consider the following calculations.
Consider two quantum systems described by Hamiltonians $H_1$ and $H_2$, whose eigensystems are separately described by
$$
\hat{H}_1|\psi_n\rangle = \epsilon_n|\psi_n\rangle,~~~~~\hat{H}_2|\phi_n\rangle = \mu_n|\phi\rangle.
$$
Let's consider case 1 first, in which the combined Hamiltonian of the combined system happens to be
\begin{equation}
\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\tag{1}
\end{equation}
Then, the eigenstates of this operator are the the product of the eigenstates of the individual Hamiltonians, and the eigenvalues are sums of the individual eigenvalues, which we can show by computing
$$
\left(\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2\right)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right)
=(\epsilon_n+\mu_m)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
(The details involve just using linearity.)
Now, given an arbitrary unentangled initial state $|\Psi(0)\rangle$, it can be written as the product of vectors expanded in the individual energy eigenbases as
$$
|\Psi(0)\rangle
=
\left(\sum_{n}a_{n}|\psi_n\rangle\right)\otimes\left(
\sum_{m}b_n|\phi_m\rangle\right)
=
\sum_{nm}a_nb_m
\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
We can then get the full time-dependence by attaching the exponential factors to the eigenvectors in the usual way, yielding
$$
|\Psi(t)\rangle
=
\sum_{nm}e^{-i(\epsilon_n+\mu_m)t/\hbar}a_nb_m
\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
Crucially, this state factors as
$$
|\Psi(t)\rangle=\left(\sum_{nm}
e^{-i\epsilon_nt/\hbar}
a_n|\psi_n\rangle\right)\otimes\left(
\sum_{m}
e^{-i\mu_mt/\hbar}
b_m|\phi_m\rangle\right),
$$
and so if the system started unentangled, it remains unentangled. This is purely a consequence of the fact that the Hamiltonian is a sum of single-system Hamiltonians because this is what leads to the eigenenergies being sums of the individual ones, which allows us to factor the exponential.
Now, to see that that doesn't work in the case of a Hamiltonian of the second form, given by
$$
\hat{H}_1\otimes\hat{H}_2\tag{2},
$$
we first note that that the product of eigenvectors is still an eigenvector, but the eigenvalues are now products of the individual eigenvalues, i.e.,
$$
\left(\hat{H}_1\otimes\hat{H}_2\right)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right)
=(\epsilon_n\mu_m)\left(|\psi_n\rangle\otimes|\phi_m\rangle\right).
$$
If we again start with the initially unentangled state shown above, then the full time-dependent state is given by
$$
|\Psi(t)\rangle
=
\sum_{nm}e^{-i(\epsilon_n\mu_m)t/\hbar}a_nb_m
\left(|\psi_n\rangle\otimes|\phi_m\rangle\right),
$$
which can no longer be factored in general!
This can also be seen by exponentiating the Hamiltonians directly to get the unitary time-evolution operators. For Hamiltonians of the form (1), the time-evolution operator is
$$
U_1(t) = \exp\left(
-\frac{it}{\hbar}
(\hat{H}_1\otimes\hat{I}_2+\hat{I}_1\otimes\hat{H}_2)
\right)
=
\exp\left(
-\frac{it}{\hbar}
\hat{H}_1\otimes\hat{I}_2
\right)
\exp\left(
-\frac{it}{\hbar}
\hat{I}_1\otimes\hat{H}_2
\right),
$$
which is allowed because the two operators commute with each other. Furthermore, it is relatively straight-forward to show that this can be written as
$$
U_1(t) =
\left(
\exp\left(
-\frac{it}{\hbar}
\hat{H}_1
\right)
\otimes\hat{I}_2\right)
\left(\hat{I}_1\otimes \exp\left(
-\frac{it}{\hbar}
\hat{H}_2
\right)\right)
=
\exp\left(
-\frac{it}{\hbar}
\hat{H}_1
\right)
\otimes
\exp\left(
-\frac{it}{\hbar}
\hat{H}_2
\right).
$$
Thus, this time-evolution operator can then be seen to "conserve unentangled-ness". The other one does not factor in that same way.
Best Answer
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.