I know we can distribute Hermitian conjugation and 'Inverses' w.r.t the Kronecker product, i.e:
$$(A\otimes B)^{-1}=(A\otimes B)^{\dagger} \leftrightarrow(A^{-1}\otimes B^{-1})=(A^{\dagger}\otimes B^{\dagger}).$$
but, does this imply $A^{-1}=A^{\dagger}$ and $B^{-1}=B^{\dagger}$? That is my question.
Take $A = 2I$ and $B=\frac 12I$ with $I$ being the identity matrix. Then $$ A\otimes B= I\implies (A\otimes B)^{-1} = I = (A\otimes B)^{\dagger} $$ but $A^{-1} \ne A = A^{\dagger}$
In general, if $X\otimes Y\ne 0$, you have that $$ X\otimes Y = M\otimes N \iff X=cM,\,\, Y=c^{-1}N $$ with $c\ne 0$. In your case, you can say that $$ A^{-1}=cA^\dagger,\,\, B^{-1}=c^{-1}B^{\dagger} $$ for a common $c$.