I see that the answer of MTyson requires making an additional
assumption, i.e., that $A\Omega$ and $B\Omega$ are diagonalizable, in
addition to the conditions described in the question. It is possible
that the conditions described in the question are too generic, hence they does not lead to any interesting form for $A$ and $B$. Here, I discuss a
different additional condition, that is more relevant for my
interests. However, I thank the other users who posted interesting
answers and suggestions!

The additional condition is that one of the two matrices, $A$ or $B$,
is positive definite. Moreover, I will assume that both $A$ and $B$
are invertible; actually, the case of singular matrices can be easily
handled in a similar (but longer) way and its discussion does not add
too much.

First, I follow the suggestion of Richard and I write $\Omega$ as:

$$ \Omega =
\begin{bmatrix}
0 & 1 & & & \cdots \\
-1 & 0 & & \\
& & 0 & 1 &\\
& & -1 & 0 & \\
\vdots & & & & \ddots
\end{bmatrix}.
$$

I claim that the two matrices $A$ and $B$ can be written as diagonal
block matrices:
\begin{gather*}
A =
\begin{bmatrix}
\lambda_1 C_1 & & & \\
& \lambda_2 C_2 & & \\
& & \lambda_3 C_3 & \\
& & & \ddots
\end{bmatrix}
\\
B =
\begin{bmatrix}
C_1 & & & \\
& C_2 & & \\
& & C_3 & \\
& & & \ddots
\end{bmatrix},
\end{gather*}
where $C_i$ are symmetric square $2n_i\times 2n_i$ matrices, and $\lambda_i$ and
$2n_i$ are the eigenvalues and their multiplicities, respectively, of
the matrix $P=B^{-1}A$. Btw, it is important to notice that $P$ is diagonalizable and all the eigenvalues have even (multiple of 2) multiplicity.

*Proof*

I call $P=B^{-1}A$. From the fact that one of the two $A$ and $B$ is positive definite, $P$ is diagonalizable.

I define $Q=-B^{-1}A\Omega$. It can be easily shown that $Q$ is antisymmetric, $Q=-Q^T$. Thus:
$$ P = Q\Omega = B^{-1}A $$
is both the product of two antisymmetric matrices ($Q$ and $\Omega$) and two symmetric matrices ($B^{-1}$ and $A$).
This ensures that the eigenvalues, $\lambda_i$, have even multiplicity, $2n_i$.

It can easily be shown that:
$$ P \Omega P^T \Omega = \Omega P^T \Omega P $$
This condition is called $\phi_J$-normality in the paper "On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation" cited by Carlo Beenakker. Together with the condition that $P$ is diagonalizable, from the same paper we conclude that $P$ is symplectically similar to a diagonal matrix:
$$ P = M^{-1} D M $$
where $M$ is symplectic.

This can be rewritten as:
$$ \tilde{B}^{-1}\tilde{A} = D $$
where $\tilde{A}=M^T A M$ and $\tilde{B}=M^T B M$.

$$ \tilde{A} = \tilde{B} D.$$

It can be noticed that $\tilde{A}$ and $\tilde{B}$ are symmetric. Thus:

$$ \tilde{B} D = D \tilde{B}.$$

Since $\tilde{B}$ commutes with a diagonal matrix, and is symmetric, it has the
block structure of the thesis. $\tilde{A}$ is then calculated as $ \tilde{A} = \tilde{B} D$.

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