First of all, there is a projection $p \in M$ of finite trace such that $N \simeq pMp$, and its equivalent class depends at most on $tr(p)$. Because $G$ is countable and $M$ a ${\rm II}_{\infty}$ factor, the measure $\mu$ must be infinite, then we can choose a subspace $Y \subset X$ with $\mu(Y)=tr(p)$, and assume that $$N = 1_Y (A \rtimes_{\alpha} G) 1_Y,$$ with $1_Y$ the indicator function of $Y$.

Let's assume$^1$ that $A$ is a Cartan subalgebra of $M$.

By the work of Feldman and Moore Ergodic equivalence relations, cohomology, and von Neumann algebras, a von Neumann algebra has a Cartan subalgebra$^2$ iff it is of the form $M(R,s)$, i.e. generated by an equivalence relation $R$ (with all equivalent classes countable) on a standard Borel space $X$ (up to a cocycle twist $s$). The Cartan subalgebra is then the "diagonal", i.e. $L^{\infty}(X)$

Now let $Y$ be a subspace of $X$, then $1_Y M(R,s) 1_Y = M(R_{|Y},s)$, so has a Cartan subalgebra equals to the diagonal $L^{\infty}(Y)$.

*Warning*: The compression $pMp$ is not necessarily a group measure space construction, because even if the equivalent relation $R_{|Y}$ is always of the form $R_H$, the action of $H$ on $Y$ is not necessarily free, whereas freeness is used to show that $M(R_H,1)$ is a group measure space construction. Moreover, the restriction $R_{|Y}$ of an ergodic$^3$ equivalent relation $R$ is not necessarily ergodic. Now, $pMp$ is a factor, so if $R_{|Y} = R_H$ with $H$ acting freely then $R_{|Y}$ must be ergodic.

*Acknowledgment*: Thanks to Jesse Peterson for his help.

$^1$it is true in the ${\rm II}_{1}$ case, but I did not check if it is always true in the ${\rm II}_{\infty}$ case.

$^2$the general case requires the existence of a faithful normal conditional expectation $E: M \to A$.

$^3$every $R$-saturated Borel set has measure $0$ or $1$.