Since $A^2 = \mathrm{tr}(A)\,A - \det(A)\,I_2$, and $\det(A)\not=0$ for $A\in M$, the leaves of your foliation are the same as the leaves of the foliation determined by the vector fields $X(A) = A$ and $Z(A) = I_2$, namely, the leaf through $A$ is the connected component containing $A$ of the (open) set $\Pi^*_A$ consisting of the matrices in the $2$-plane $\Pi_A\subset M_2(\mathbb{R})$ spanned by $I_2$ and $A$ that are not multiples of the identity and that are invertible.

When $A$ has real distinct eigenvalues, the open set $\Pi^*_A$ has $6$ components, the complement of $3$ lines in $\Pi_A$ that intersect at $0\in M_2(\mathbb{R})$. When $A$ has a double eigenvalue, $\Pi^*_A$ has $4$ components, the complement of two lines through $0\in \Pi_A$. When $A$ has no real eigenvalues, $\Pi^*_A$ has $2$ components, the compliment of a single line in $\Pi_A$.

Thus, all of the leaves are simply connected, so there is no possibility for nontrivial holonomy.

The leaf space, which is formally of dimension $2$, is not Hausdorff, precisely because of the leaves containing an $A$ that has a double eigenvalue. There is a smooth submersion $\pi:M\to S^2$ such that $\pi(A)$ is the oriented ray in the $3$-dimensional space of traceless $2$-by-$2$ matrices that consists of the positive multiples of $A -\tfrac12\mathrm{tr}(A)\, I_2$. The fibers of $\pi$ are 'half-planes' in the submanifolds $\Pi^*_A$, but the leaves of the foliation are connected components of the $\Pi^*_A$, so the number of leaves in the $\pi$-preimage of a point of $S^2$ varies between $1$ and $3$. It is at the places where the number of leaves in the preimage equals 2 (a union of two circles in $S^2$) that the natural local charts provided by localizing $\pi$ fail to separate points.

As a specific example, let $A_+ = \begin{pmatrix}1&1\\0&1\end{pmatrix}$ and let $A_- = \begin{pmatrix}-1&1\\0&-1\end{pmatrix}$. Then it is not hard to show that the leaves through these two points cannot be separated by any function that is smooth on the leaf space endowed with the quotient smooth structure.

Finally, if you just give $M$ the standard flat metric that it inherits as an open set in $M_2(\mathbb{R})\simeq \mathbb{R}^4$, then the leaves will be totally geodesic.