All matrix rings are over the integers, colimits are colimits of $\mathsf{Rng}$s and $\otimes$ means $\otimes_{\mathbb{Z}}$.
My guess is that yes, indeed, and my reasoning is as follows: since $M_\infty = \mathrm{colim} M_n$, then
$$ M_\infty \otimes M_\infty \simeq \mathrm{colim}_{(i,j)} M_i \otimes M_j \simeq \mathrm{colim}_{(i,j)} M_{ij} $$
and similarly
$$ M_n \otimes M_\infty \simeq \mathrm{colim} M_{in}. $$
Both of these are isomorphic to $M_\infty$ if the following is correct,
Proposition: let $f : (\Lambda,\leq) \to (\mathbb{N},\leq)$ be a cofinal order-preserving function. Then $\mathrm{colim}_{\alpha \in \Lambda} M_{f(\alpha)} \simeq M_\infty$.
Proof. For each $\alpha \in \Lambda$, the identity maps of $M_{f(\alpha)}$ composed with inclusion to $M_\infty$ assemble into a map
$$ \gamma : \mathrm{colim}_{\alpha \in \Lambda} M_{f(\alpha)} \to M_\infty $$
Picking a matrix in $M_\infty$, by cofinality it is contained in some $M_{f(\alpha)}$ and so it is in the image of $\gamma$. Likewise, if an element in the colimit is given by a matrix in $M_{f(\alpha)}$, its image in $M_\infty$ coincides with the image via the monomorphism $M_{f(\alpha)} \hookrightarrow M_\infty$ and thus $\gamma$ is injective. $\square$
Is this proof correct? Is there a more categorical proof? That is, can we deduce that $\gamma$ is an iso without appealing to bijectivity? Also, there seems to be some general fact about diagrams indexed by (filtered?) posets hiding somewhere.
As a side question (which I think is a bit to narrow not to include here),
Is the morphism $M_\infty \to M_\infty \otimes M_\infty$ given by tensoring the inclusion $\mathbb{Z}\ni k \mapsto k \cdot e_{11} \in M_\infty$ with $M_\infty$ an iso?