Let $R$ be a commutative ring, let $S\subset R$ be multiplicative and let $M$ be an $R$-module. Is the following true?
The localization map $M\to S^{-1}M$ is an isomorphism if and only if for all $R$-modules $W$ such that $S^{-1}W=0$, all maps $W\to M$ are zero.
The direction $\Rightarrow$ is easy: it follows from functoriality of localization. But I can't prove or disprove $\Leftarrow$. Note that the condition $S^{-1}W=0$ implies that all elements of $W$ are annihilated by some element of $S$.
I think $\mathbb{Z} \rightarrow \mathbb{Q}$ (so $R = M = \mathbb{Z}$ and $S = \mathbb{Z} \setminus \{0\}$) provides a counterexample. Given an abelian group $W$, then $S^{-1} W = 0$ iff $W$ is torsion, i.e., every element has finite order. Since $0$ is only element of $\mathbb{Z}$ of finite order, given a homomorphism $W \to \mathbb{Z}$, then every element must map to $0$. Thus the hypothesis on the righthand side is satisfied, but the localization map $\mathbb{Z} \to \mathbb{Q}$ is not an isomorphism.
However, one can show the localization map is injective given the hypothesis on the righthand side. Letting $\varphi: M \to S^{-1} M$ be the localization map, take $W = \ker(\varphi)$. Given $m \in \ker(\varphi)$, then $\frac{ms}{s} = \frac{0}{s}$, so there exists $t \in S$ such that $$ 0 = t(ms^2 - 0\cdot s) = s^2 t m \, . $$ Thus $S^{-1}W = 0$. By the hypothesis, then the inclusion map $\ker(\varphi) \hookrightarrow M$ must be the zero map, so $\ker(\varphi) = 0$.