The answer to your question, as I now understand it, is no. In particular, we can construct a matrix of you particular pattern with a positive determinant that fails to be positive definite.
In particular, consider the matrix
$$
M =
\pmatrix{
1&-1&-1&0&0&0\\
-1&1&-1&0&0&0\\
-1&-1&1&0&0&0\\
0&0&0&1&-1&-1\\
0&0&0&-1&1&-1\\
0&0&0&-1&-1&1\\
}
$$
which has eigenvalues $-1-1,2,2,2,2$
Your guess of $S=R$ is right if $A$ is positive definite and $B$ is symmetric.
Eigen decomposition of $A$: $A = Q \Lambda Q^T$ with orthogonal $Q$ and diagonal $\Lambda>0$.
We obtain $A = P\,P^T$ with $P:=Q\Lambda^{-\frac12}$.
The matrix $P^T B P$ is symmetric. Therefore, it admits an eigenvalue decomposition $D = V^T P^T B P V$ with diagonal $D$ and orthogonal $V$.
We have $1 = V^T \underbrace{P^T A P}_{=1} V$ as required.
Thus $S:= PV = Q\Lambda^{-\frac12}V$ is the first wanted transformation matrix.
Now we test whether we can use $R:=S$ for $R^{-1} A^{-1} B R \overset{?}{=} D$.
\begin{align*}
S^{-1}A^{-1}BS &= \underbrace{V^T \Lambda^{\frac12} Q^T}_{S^{-1}} \underbrace{Q\Lambda^{-1} Q^T}_{A^{-1}}\, B\, \underbrace{Q \Lambda^{-\frac12} V}_{S}\\
&= V^T \underbrace{\Lambda^{-\frac12}Q^T}_{P^{T}} B \underbrace{Q \Lambda^{-\frac12}}_{P} V\\
&= D
\end{align*}
Best Answer
How large is your matrix? Perhaps you can post it here. I would venture to guess that you may be able to apply the Gershgorin Circle Theorem. This is noted in Calle's answer in the question you linked to.