First approach: If $\{\lambda_1,\dots,\lambda_m\}$ are the eigenvalues of $A$ and $\{\nu_1,\dots,\nu_n\}$ those of $B$, then the eigenvalues of $A\otimes B$ are $\lambda_j\cdot\mu_k,1\leq j\leq m,1\leq k\leq n$.
We assume that the respective dimensions of $A$ and $B$ are $m$ and $n$. If $v$ is an eigenvector of $A$ for $\lambda_k$ and $w$ of $B$ for $\mu_j$, consider $V$ the vector of size $mn$, defined by
$$V=(v_1w_1,\dots,v_1w_n,v_2w_1,\dots,v_2w_n,\dots,v_mw_1,v_mw_n).$$
It's an eigenvector of $A\otimes B$ for the eigenvalue $\lambda_k\mu_j$. As the matrices $A$ and $B$ are diagonalizable, counting multiplicity we are sure there aren't other eigenvalues.
As $A$ and $B$ are positive definite, $\lambda_k\mu_j>0$ for all $k,j$.
Second approach: We use mix product property, that is $$(A_1A_2)\otimes (B_1B_2)=(A_1\otimes B_1)(A_2\otimes B_2).$$
Applied twice, this gives $$A\otimes B=(P_1^tD_1P_1)\otimes (P_2^tD_2P_2),$$
where $P_i$ are orthogonal and $D_i$ diagonal. This gives
$$A\otimes B=(P_1\otimes P_2)^t(D_1\otimes D_2)(P_1\otimes P_2),$$
so the problem reduces to the case $A$ and $B$ diagonal, which is easy, as the eigenvalues are positive.
What you want to prove is not necessarily true. Here is a random counterexample:
$$
D=\left[\begin{array}{c|c}A&C\\ \hline C^T&B\end{array}\right]
=\left[\begin{array}{rr|rr}
2&-3&1&0\\
2&2&0&-1\\
\hline
1&0&2&1\\
0&-1&-2&2
\end{array}\right].
$$
According to your definition, a matrix is "positive definite" if and only if its symmetric part is positive definite in the conventional sense. Now, WolframAlpha reckons that the eigenvalues of
$$
D+D^T=\left[\begin{array}{rr|rr}
4&-1&2&0\\
-1&4&0&-2\\
\hline
2&0&4&-1\\
0&-2&-1&4
\end{array}\right]
$$
are $4-\sqrt{5}$ and $4+\sqrt{5}$ (each of multiplicity 2). Hence $D$ is "positive definite". WolframAlpha also reckons that $\det(D)=61$. However,
$$\det(A)\det(B)=10\times6=60<\det(D).$$
Best Answer
Yep, we can say things. The two ingredients you need are the following:
Equipped with these things, you can find out more quantitative relationships between the old and new matrices.
Edit: The question has been updated to ask about relationships between the ratios of the determinants, which according to the matrix determinant lemma are the quantities $$1+v^TA^{-1}v\quad \text{and}\quad 1+v^TB^{-1}v$$ The short general answer is no. Even though we have some relationships between the spectra of $A$ and $B$, we can't in general say much about $v^TA^{-1}v$ compared to $v^TB^{-1}v$, since it may happen that $v$ is well-aligned with an eigenvector of $A^{-1}$ of large eigenvalue, and an eigenvector of $B^{-1}$ of small eigenvalue.
An extreme example meeting all conditions in the question would be to take a positive definite matrix $A$ which has large variation in its spectrum, and make $B$ be a rotated copy of $A$, i.e. obtained by conjugating with an orthogonal transformation. Then it may well happen that for some $v$ the ratio of determinants is huge, while for others it's tiny.