Here's my second attempt at an answer.

As before, I claim that in general, anything can happen. That is, the inertia of $A\circ B$ can be almost anything (i.e., can be very different from the inertia of either $A$ or $B$).

Here is a line of thought.

- Let $A$ be psd, and let $B=ee^T$ (the all ones matrix). Then, $B$ has just nonzero eigenvalue, which is $e^Te$; but inertia of $A \circ B$ is equal to inertia of $A$.
- Let $B=-ee^T$ instead, then inertia of $A\circ B$ is equal to "negative" inertia of $A$
- The above two examples suggest to me that by carefully manipulating $B$ we can make $A\circ B$ have any kind of inertia.
- To substantiate this intuition, consider the following example:

\begin{equation*}
A=\begin{bmatrix}
9 & 1 & -4\\\\
1 & 2 & -6\\\\
-4&-6 & 24
\end{bmatrix},\quad
c = \begin{bmatrix}
5\\\\
-4\\\
-1
\end{bmatrix}.
\end{equation*}

Now by setting $B=cc^T + \alpha ee^T$, and varying $\alpha$, you can construct several examples where the inertia of $A\circ B$ ranges from all positive eigenvalues to all negative eigenvalues (this examples excludes the all zeros case), while being different from the inertia of $B$ (or with suitable modification, of $A$).

Some of the above search was guided by inequalities such as $\lambda(A\circ B) \ge (\min_i a_ii)\lambda_\min(B)$. Thus, in general, I think it is easier to derive inequalities for the eigenvalues of Hadamard products, than full characterizations. The only other idea worth mentioning here is that since the Hadamard product matrix is a principal submatrix of the Kronecker product, perhaps a more fine-grained statement can still be made.