3 fixed minor error.

Assuming that $X$ and $Y$ are invertible (if not, probably a perturbation argument will yield the generalization). Then, the following choice of $M$ works, i.e., $MDM^T$ equals your original matrix with $D$ being the anti-diagonal identity matrix as desired.

(Also, note slightly different notation, I write $-Y$ instead of $Y$ as in the original question, so that both $X$ and $Y$ are positive).

A solution is given by: \begin{equation*} M = \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \end{equation*}

where

\begin{equation*} a = X^{1/2},\quad b=\frac{X^{1/2}}{2},\quad c = X^{-1/2} - (X^{-1}+Y)^{1/2},\quad d = \frac{X^{-1/2} - + (X^{-1}+Y)^{1/2}}{2}. \end{equation*}

Edit. Typo in $d$ fixed now.

To see that the above matrix provides a solution, simply verify

\begin{equation*} \begin{bmatrix} a & b\\ c & d \end{bmatrix}\begin{bmatrix} 0 & I\\ I & 0\end{bmatrix}\begin{bmatrix} a^T & c^T\\ b^T & d^T\end{bmatrix} = \begin{bmatrix} X & I\\ I & -Y\end{bmatrix}, \end{equation*} which boils down to checking the following four equations: \begin{eqnarray*} ad^T + bc^T &=& I\\ cb^T + da^T &=& I\\ ab^T + ba^T &=& X\\ cd^T + dc^T &=& -Y. \end{eqnarray*}

Assuming that $X$ and $Y$ are invertible (if not, probably a perturbation argument will yield the generalization). Then, the following choice of $M$ works, i.e., $MDM^T$ equals your original matrix with $D$ being the anti-diagonal identity matrix as desired.

(Also, note slightly different notation, I write $-Y$ instead of $Y$ as in the original question, so that both $X$ and $Y$ are positive).

A solution is given by: \begin{equation*} M = \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \end{equation*}

where

I arrived at this solution after

To see that the above matrix provides a somewhat circuitous derivationsolution, but can provide simply verify

\begin{equation*} \begin{bmatrix} a & b\\ c & d \end{bmatrix}\begin{bmatrix} 0 & I\\ I & 0\end{bmatrix}\begin{bmatrix} a^T & c^T\\ b^T & d^T\end{bmatrix} = \begin{bmatrix} X & I\\ I & -Y\end{bmatrix}, \end{equation*} which boils down to checking the details if needed. following four equations: \begin{eqnarray*} ad^T + bc^T &=& I\\ cb^T + da^T &=& I\\ ab^T + ba^T &=& X\\ cd^T + dc^T &=& -Y. \end{eqnarray*}

1

Assuming that $X$ and $Y$ are invertible (if not, probably a perturbation argument will yield the generalization). Then, the following choice of $M$ works, i.e., $MDM^T$ equals your original matrix with $D$ being the anti-diagonal identity matrix as desired.

\begin{equation*} M = \begin{bmatrix} a & b\\ c & d \end{bmatrix}, \end{equation*}

where