Suppose $A$ is ring and $A_S$ denotes localized ring when $S$ is a multiplicatively closed subset of $A$ containing $1$ and $\phi: A\to A_S$ is the canonical map $r\mapsto \frac{r}{1}$ where $\frac{a}{b}=$ equivalance class containing $(a,b)$. I was able to prove the following.
$A$ is an integral domain $\implies$ $\phi$ is an embedding.
Proof: $\phi(x)=\phi(y)\implies\frac{x}{1}=\frac{y}{1}\implies\exists s\in S:s(x-y)=0$. By definition multiplicatively closed sets do not contain $0$ so $s\neq0$. As $A$ is an integral domain $x-y=0\implies x=y$. Therefore, $\phi$ is one-to-one.
But I'm unable to prove the converse. Is it true at all?
Suppose $s$ and $t$ are nonzero elements of $A$ with $st=0$. Consider $S=\{1,s,s^2,\ldots\}$. Then $t/1=0$ in $A_S$.