Classical Hamiltonians
By the spectral theorem, for every Hamiltonian there exists a basis in which it is diagonal. Thus, it is not correct to say that diagonal Hamiltonians are classical since this would apply to all Hamiltonians.
A Hamiltonian $H$ which is diagonal in a product basis $\mathcal{B}$ is sometimes described as classical, because the evolution of any initial state in $\mathcal{B}$ driven by such $H$ does not give rise to quantum effects such as superposition and entanglement. Moreover, a measurement in $\mathcal{B}$ at any point during the evolution yields a single, deterministic result and does not entail wavefunction collapse.
Note that the designation of certain bases as product bases and of certain Hamiltonians as classical depends on the choice of partitioning of a composite system into subsystems.
Commuting tensor factors
The following characterization of classical Hamiltonians is useful.
Theorem Let $H$ be a Hermitian operator on a two-body Hilbert space $\mathcal{H}_A\otimes\mathcal{H}_B$. Then $H$ is diagonal in a product basis if and only if it can be written as $H=\sum_{i=0}^{k-1}A_i\otimes B_i$ for some integer $k$ and Hermitian operators $A_i$ and $B_i$ such that $A_i$ commutes with $A_j$ for every $i,j=0,\dots,k-1$ and $B_i$ commutes with $B_j$ for every $i,j=0,\dots,k-1$.
Proof Suppose $H$ is diagonal in the product basis $|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$ with $\alpha=0,\dots,m-1$ and $\beta=0,\dots,n-1$ where $m=\dim\mathcal{H}_A$ and $n=\dim\mathcal{H}_B$. The spectral theorem guarantees that $|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$ are orthogonal. Therefore, $|\psi_\alpha\rangle$ are orthogonal and $|\phi_\beta\rangle$ are orthogonal. Moreover, we can choose both bases to be orthonormal. Then, the eigendecomposition of $H$ is
$$
H=\sum_{\alpha=0}^{m-1}\sum_{\beta=0}^{n-1}\lambda_{\alpha\beta}|\psi_\alpha\rangle\langle\psi_\alpha|\otimes|\phi_\beta\rangle\langle\phi_\beta|
$$
where $\lambda_{\alpha\beta}$ is the eigenvalue of $H$ associated with eigenvector $|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$. Let $k=mn$. Any $i=0,\dots,k-1$ can be written as $i=\alpha n + \beta$ for unique $\alpha=0,\dots,m-1$ and $\beta=0,\dots,n-1$. Define $A_{\alpha n + \beta}=\lambda_{\alpha\beta}|\psi_\alpha\rangle\langle\psi_\alpha|$ and $B_{\alpha n + \beta}=|\phi_\beta\rangle\langle\phi_\beta|$. Then
$$
\begin{align}
H &= \sum_{\alpha=0}^{m-1}\sum_{\beta=0}^{n-1}\lambda_{\alpha\beta}|\psi_\alpha\rangle\langle\psi_\alpha|\otimes|\phi_\beta\rangle\langle\phi_\beta| \\
&=\sum_{\alpha=0}^{m-1}\sum_{\beta=0}^{n-1}A_{\alpha n+\beta}\otimes B_{\alpha n+\beta} \\
&= \sum_{i=0}^{k-1}A_i\otimes B_i\tag1
\end{align}
$$
and $A_i$ commutes with $A_j$ for every $i,j=0,\dots,k-1$ and $B_i$ commutes with $B_j$ for every $i,j=0,\dots,k-1$.
Conversely, suppose that $H$ can be written as $H = \sum_{i=0}^{k-1}A_i\otimes B_i$ with Hermitian operators $A_i$ and $A_j$ commuting for every $i,j=0,\dots,k-1$ and Hermitian operators $B_i$ and $B_j$ commuting for every $i,j=0,\dots,k-1$. The operators $A_i$ share a common eigenbasis, as do operators $B_i$. Let $|\psi_\alpha\rangle$ with $\alpha=0,\dots,m-1$ denote the shared eigenbasis of operators $A_i$ and $|\phi_\beta\rangle$ with $\beta=0,\dots,n-1$ the shared eigenbasis of operators $B_i$. We can write $A_i=\sum_{\alpha=0}^{m-1} \mu_{i,\alpha}|\psi_\alpha\rangle\langle\psi_\alpha|$ and $B_i=\sum_{\alpha=0}^{n-1}\nu_{i,\beta}|\phi_\beta\rangle\langle\phi_\beta|$ and
$$
\begin{align}
H &= \sum_{i=0}^{k-1}A_i\otimes B_i \\
&= \sum_{i=0}^{k-1}\left( \sum_{\alpha=0}^{m-1} \mu_{i,\alpha}|\psi_\alpha\rangle\langle\psi_\alpha|\right)\otimes \left(\sum_{\beta=0}^{n-1}\nu_{i,\beta}|\phi_\beta\rangle\langle\phi_\beta|\right) \\
&= \sum_{\alpha=0}^{m-1} \sum_{\beta=0}^{n-1} \left(\sum_{i=0}^{k-1} \mu_{i,\alpha}\nu_{i,\beta}\right) |\psi_\alpha\rangle\langle\psi_\alpha|\otimes |\phi_\beta\rangle\langle\phi_\beta| \\
&= \sum_{\alpha=0}^{m-1} \sum_{\beta=0}^{n-1} \lambda_{\alpha\beta} |\psi_\alpha\rangle\langle\psi_\alpha|\otimes |\phi_\beta\rangle\langle\phi_\beta| \tag2
\end{align}
$$
where we defined $\lambda_{\alpha\beta} = \sum_{i=0}^{k-1} \mu_{i,\alpha}\nu_{i,\beta}$. Equation $(2)$ exhibits eigendecomposition of $H$ in the product basis $|\psi_\alpha\rangle\otimes|\phi_\beta\rangle$ completing the proof. $\square$
Remark 1: The theorem applies to many-body operators as well. This can be seen by suitable generalization of the proof above.
Remark 2: Note that in the above characterization of classical Hamiltonians it is key that tensor factors on each subsystem commute pairwise across all terms. It is insufficient for the whole terms on the composite system to commute pairwise. An example of a non-classical Hamiltonian with commuting terms is
$$
\begin{align}
H &= -\frac12\sigma_1^X \sigma_2^X - \frac12\sigma_1^Z \sigma_2^Z \\
&= -|\phi^+\rangle\langle\phi^+| + |\psi^-\rangle\langle\psi^-|
\end{align}
$$
whose non-degenerate ground state is $|\phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$. Another related example of a non-classical Hamiltonian with commuting terms is the Toric Code.
(Thanks to @NorbertSchuch who pointed out an error in an earlier version of this section and provided the above counterexample.)
The theorem has two consequences for the Hamiltonians in the question. Namely, it implies that the first Hamiltonian is classical and it justifies the addition of a transversal field in the second one as a valid attempt to make it non-classical. Its non-classical character can be confirmed by diagonalization.
Terminology
Commutation is a relation between two operators, not a property of an operator. However, the expression "commuting Hamiltonian" is used sometimes as a shortcut for "Hamiltonian with pairwise commuting terms".
