Let $R$ be a ring, $S \subset R$ a multiplicatively closed subset and $M$ and $R$-module.
There is a bilinear map $S^{-1}R \times M \to S^{-1}M$ given by $(\frac{r}{s}, m) \mapsto \frac{rm}{s}$ giving rise to an $R$-linear map $$f: S^{-1}R \otimes_R M \to S^{-1}M$$ defined by $\frac{r}{s} \otimes m \mapsto \frac{rm}{s}$. To show that localisation is an example of an extension of scalars, I would like to show that $f$ is an isomorphism.
For surjectivity just take $r=1$.
I don't know how to show that this map is injective. Let $\frac{r}{s} \otimes m \in \ker f$. Then $\frac{rm}{s} = \frac{0}{1}$ and there exists $s' \in S$ such that $rms'=0$, but now what?
Doesn't $f$ actually just tell us about the structure of both sets as $R$-modules? I believe that to show they are isomorphic as $S^{-1}R$-modules we would need to check that the action of $S^{-1}R$ is the same on both sides? Perhaps this is trivial and omitted.
For 1) You observed that $r/s \otimes m \in \ker{f}$ implies $s'rm = 0$ for some $s' \in S$. Then use the defining laws of tensor products to manipulate the expression $r/s \otimes m$. We have $r/s \otimes m = 1/s \otimes rm = \frac{s'}{ss'} \otimes rm = \frac{1}{ss'} \otimes s'rm = \frac{1}{ss'} \otimes 0 = 0$, as desired.
For 2) Suppose that two $S^{-1}R$ modules $M, N$ are isomorphic as $R$-modules. So we have an $R$-linear map $f: M \rightarrow N$ that is an isomorphism. In fact, $f$ is necessarily also $S^{-1}R$ linear. To see this, consider the expression $r/s f(m) - f(rm/s)$. Multiply by $s$ and use $R$-linearity of $f$ to see that this expression is $0$ in $N$. So actually $f: M \rightarrow N$ is an isomorphism of $S^{-1}R$ modules! In conclusion, if two $S^{-1}R$ modules $M, N$ are isomorphic as $R$-modules, then they are isomorphic as $S^{-1}R$-modules.
PS) Perhaps it's worth noting that the argument in 2) generalizes from localizations to epimorphisms of rings, i.e. if $R \rightarrow T$ is an epimorphism of commutative rings, and $M, N$ are two $T$-modules, then an $R$-linear map $f: M \rightarrow N$ is already $T$-linear, hence $M, N$ are isomorphic as $T$ modules if they are isomorphic as $R$-modules.