I know that the product $AB$ of a Hermitian and positive definite matrix $A$ and a Hermitian matrix $B$ is itself Hermitian.
From simulations (where $A$ is diagonal, but I don't think this matters), I have a suspicion that the signature (number of positive/negative eigenvalues) of $AB$ is the same as $B$. I think Sylvester's law of inertia is applicable, but I don't know how to view $A$ as a transformation under which the signature is invariant. Also, I don't know anything about quadratic forms, so please bear with me.
I'd like to use this result in my (physics) bachelor's thesis, so any help is greatly appreciated!
No, that isn't true. Counterexample: $\pmatrix{1&0\\ 0&2}\pmatrix{1&1\\ 1&1}=\pmatrix{1&1\\ 2&2}$ is not Hermitian.
However, in case $AB$ is Hermitian, we get $AB=(AB)^\ast=B^\ast A^\ast=BA$. Hence $B$ also commutes with $A^{1/2}$ and $AB=A^{1/2}BA^{1/2}$. Therefore, by Sylvester's law of inertia, $AB$ and $B$ have identical signature.