Find the right positive definite block matrix Let $0<a<b$, and let $A, B \in \mathbb{R}^{n \times n}$ be two positive definite symmetric matrices (that do not commute).

My question is: are the matrices $M, P \in \mathbb{R}^{2n \times 2n}$ defined by
  \begin{align*}
M = \begin{bmatrix}
bABA & aAB\\
aBA & bBAB
\end{bmatrix},\qquad P = \begin{bmatrix}
bA & aAB\\
aBA & bB
\end{bmatrix}
\end{align*}
  positive definite or not? (In the sense $x^\intercal M x >0$ for any nonzero $x \in \mathbb{R}^{2n}$.)

Why this question: I am trying to find a positive definite symmetric matrix $Q \in \mathbb{R}^{2n \times 2n}$ such that the product
\begin{align*}
Q \begin{bmatrix}
0 & A^{-1}\\
B^{-1} & 0
\end{bmatrix}
\end{align*}
is a symmetric matrix. My candidates at the moment are M and P, which are such that
\begin{align*}
M \begin{bmatrix}
0 & A^{-1}\\
B^{-1} & 0
\end{bmatrix} = \begin{bmatrix}
aA & bAB\\
bBA & aB
\end{bmatrix}, \qquad P \begin{bmatrix}
0 & A^{-1}\\
B^{-1} & 0
\end{bmatrix} = \begin{bmatrix}
aA & bI_n\\
bI_n & aB
\end{bmatrix}
\end{align*}
but I don't know if they are positive definite.
What I tried: I tried to find an invertible matrix $K$ such that the product $K^\intercal P K$ (or $K^\intercal M K$ ) is easier to study, since $P$ is positive definite if and only if $K^\intercal P K$ is. For example, for
\begin{align*}
K = \begin{bmatrix}
A^{-1} & A^{-1}\\
B^{-1} & B^{-1}
\end{bmatrix}
\end{align*}
we have
\begin{align*}
K^\intercal P K = \begin{bmatrix}
bA^{-1} + 2a I_n + bB^{-1} & bA^{-1} - bB^{-1} \\
bA^{-1} - bB^{-1} & bA^{-1} - 2a I_n + bB^{-1}
\end{bmatrix}.
\end{align*}
But I don't know how to proceed for this matrix either.

First of all, we must assume that $b>0$ for either matrix to be positive definite.
For $M$, we compute the Schur complement
$$
bABA - \frac{a^2}{b}(AB)(BAB)^{-1}(BA)=
bABA - \frac{a^2}{b}A.
$$
Since $bABA$ is positive definite, $M$ will be positive definite if and only if $bABA - \frac{a^2}{b}A$ is positive definite.  With the Loewner order, we can write this condition as 
$$
bABA - \frac{a^2}{b}A > 0 \iff
bABA > \frac{a^2}{b}A \iff \\
ABA > \frac{a^2}{b^2} A \iff
A^{-1/2}(ABA)A^{-1/2} > A^{-1/2}[\frac{a^2}{b^2} A]A^{-1/2} \iff\\
A^{1/2}BA^{1/2} > \frac{a^2}{b^2}I.
$$
That is, $M$ will be positive definite if and only if the eigenvalues of $A^{1/2}BA^{1/2}$ are greater than $a^2/b^2$.  We see that the matrix $AB$ is similar since
$$
AB = A^{1/2}(A^{1/2}BA^{1/2})A^{-1/2},
$$
so $M$ will be positive definite if and only if the eigenvalues of $AB$ are greater than $a^2/b^2$.
For $P$, we compute the Schur complement to be
$$
bA - \frac{a^2}{b} ABA
$$
and a similar analysis can be applied.  We find that $P$ will be positive definite if and only if either $a=0$ or the eigenvalues of $AB$ are less than $b^2/a^2$.
In summary, $M$ and $P$ will be positive definite if and only if $a=0$ or the eigenvalues of $AB$ lie inside the interval $(a^2/b^2,b^2/a^2)$.

A potentially helpful insight: we can reduce your original problem to a possibly simpler case by considering a congruent matrix. For instance:
$$
\pmatrix{A^{1/2}\\&B^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&B^{1/2}} = 
\pmatrix{0&A^{-1/2}B^{1/2}\\B^{-1/2}A^{1/2} & 0}\\
%
\pmatrix{A^{1/2}\\&A^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&A^{1/2}} = 
\pmatrix{0&I\\A^{1/2}B^{-1}A^{1/2} & 0}
$$
