Let $M_1, \dots, M_n$ be $A$-modules. Describe an isomorphism $$S^{-1}(M_1 \times \dots \times M_n) \to S^{-1}(M_1) \times \dots \times S^{-1}(M_n).$$
This is essentially showing that the localization commutes with finite products. If we define $$\begin{align*} S^{-1}(M_1 \times \dots \times M_n) &\to S^{-1}(M_1) \times \dots \times S^{-1}(M_n) \\\\ \frac{(m_1, \dots, m_n)}{(s_1,\dots,s_n)} &\mapsto (m_1/s_1, \dots, m_n/s_n)\end{align*}$$
and $$\begin{align*} S^{-1}(M_1) \times \dots \times S^{-1}(M_n) &\to S^{-1}(M_1 \times \dots \times M_n) \\\\ (m_1/s_1, \dots, m_n/s_n) &\mapsto \frac{(m_1, \dots, m_n)}{(s_1,\dots,s_n)}\end{align*}$$
will these give us an isomorphism or do we need the universal property here?
I'm not sure why you have ordered tuples of $s_i$. I think what you mean to write is $(m_1, \dots, 0)/s_1 + (0, m_1, \dots, 0)/s_2 + \cdots + (0, \dots, m_n)/s_n$, since elements of $S^{-1}(M_1 \times \cdots \times M_n)$ will not all be simple products $r/s (m_1, \dots, m_n)$. This is essentially the fact that elements of the tensor product are not all simple tensors, but rather sums of simple tensors.
You could do this directly with universal properties, but I'll suggest a more indirect way using the splitting lemma. Consider the exact sequence
$$0 \longrightarrow M_1 \longrightarrow M_1 \times M_2 \longrightarrow M_2 \longrightarrow 0$$
where we have a splitting map $M_2 \to M_1 \times M_2$ given by $m_2 \mapsto (0, m_2)$. We can then apply the exact functor $S^{-1}(-)$ to get
$$0 \longrightarrow S^{-1}M_1 \longrightarrow S^{-1}(M_1 \times M_2) \longrightarrow S^{-1}M_2 \longrightarrow 0$$
and notice that by the definition of a functor, the splitting map will be sent to another splitting map. Hence, this is a split exact sequence so we have an isomorphism $S^{-1}(M_1 \times M_2) \to S^{-1}M_1 \times S^{-1}M_2$ by the splitting lemma, which is described very concretely! In fact you'll find this isomorphism very familiar.
Alternatively, you could show the more general fact that if $A \to B$ is a ring homomorphism and $M_1, M_2$ are $A$-modules, then $B \otimes (M_1 \times M_2) \cong (B \otimes M_1) \times (B \otimes M_2)$. This is described in a lot of books, but also here.