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.
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.