The answer is yes, following from Klee's extension of Keller's theorem on homeomorphisms of infinite dimensional compact convex sets:
Klee, V. L., Some topological properties of convex sets, Trans. Am. Math. Soc. 78, 30-45 (1955). ZBL0064.10505.
Specifically, combining (1.1) with the theorem of Keller:
If $K$ is an infinite-dimensional compact convex subset of $E$, a topological vector space, such that there is a countable set $F \subseteq E^*$ separating the points of $K$, then $K$ is homeomorphic to the Hilbert cube.
(For the avoidance of doubt, I use $E^*$ to mean the space of continuous linear functionals throughout)
For $\mathcal{M}_1(S)$, consider it inside $C(S)^*$ with the weak-* topology and take $F$ to be your favourite countable norm-dense subset of $C(S)$, considered as weak-* continuous maps $C(S)^* \rightarrow \mathbb{R}$. For $\mathcal{M}_1(S) \times \mathcal{M}_1(S)$, consider it inside the product topological vector space $C(S)^* \times C(S)^*$ with the product of weak-* topologies (the underlying Banach space is the $\ell^\infty$-direct sum). Define
$$
F' = \{ \phi \circ \pi_1 \mid \phi \in F \} \cup \{ \phi \circ \pi_2 \mid \phi \in F \}
$$
where $\pi_1 : E \times F \rightarrow E$ and $\pi_2 : E \times F \rightarrow F$ are the usual projection maps. Then $F'$ is a countable subset of $(C(S)^* \times C(S)^*)^*$ separating the points of $\mathcal{M}_1(S) \times \mathcal{M}_1(S)$, so this is also homeomorphic to the Hilbert cube and therefore to $\mathcal{M}_1(S)$.