There was a quite fast answer to this question regarding invertibility of $M$, but as I discovered more properties, I thought we must be able to find stronger results.
$$M = \begin{bmatrix}I_k&A\\A^T&-I_l\end{bmatrix}$$
Experimental observations
Based on numerical experiments with random matrices $A$ I think this shall be possible to prove that we will have $|k-l|$ eigenvalues at $1$ if $k>l$ and at $-1$ if $l>k$ and the rest in pairs $(1-t, 1+t)$ where $t>0$.
And that corresponding eigenvector pairs are orthogonal to one another.
Own theoretical observations
Now by Gershgorins circle theorem the eigenvalues are in circles with the radii the sum absolute values of A rows / columns, but midpoint different ($k$ have $+1$ and $l$ have $-1$).
By the spectral theorem since $M$ is symmetric also all eigenvalues will be along the real line.
But since $A$ can be any matrix, I cannot bound the radii, and it was too long ago when I last saw techniques used to bound these radii. I have some weak memory mentioning techniques using continuity to get better Gershgorin estimates. But I don't remember how to do it. Any pointers would be helpful.
The essential case is this one
$\textbf{Proposition.}$ Assume that $k<l$ and that $A$ has full rank $k$. Let $(\sigma_i)_{i\leq k}$ be the $k$ positive singular values of $A$. Then
$spectrum(M)=\{\pm\sqrt{1+\sigma_i^2};i=1,\dots,k\}\cup \{-1,\cdots,-1; l-k\;\times\}$.
$\textbf{Proof.}$ Note that $AA^T$ is invertible and $spectrum(AA^T)=(\sigma_i^2)_i$. If $\lambda\not= -1$, then
$\det(M-\lambda I_{k+l})=\det((1-\lambda)I_k-A(-1-\lambda)^{-1}A^T)\det((-1-\lambda)I_l)=0$
IFF $\det((1-\lambda^2)I_k+AA^T)=0$.
Thus $\lambda$ is in the form $\pm\sqrt{1+\sigma_i^2}\notin[-1,1]$. These are the $2k$ eigenvalues of $M$ (with multiplicity) which are $\not= -1$. Clearly, the $l-k$ remaining eigenvalues are equal to $-1$. In particular, we can verify that $tr(M)=-(l-k)$. $\square$