Let $R$ a ring and $S$ subset of $R$ that is closed under the multiplication. We denote $$S^{-1}R=(R\times S)/_\sim,$$ where $$(r,s)\sim(r',s')\iff \exists s''\in S: s''(rs'-r's)=0.$$
This equivalence relation $\sim$ looks weird to me. I tried to apply it an an example, for example $R=\mathbb Z/6\mathbb Z$ and $S=2\mathbb Z/6\mathbb Z$. How does it will work here ? I don't see it ! May be you have an easier example ?
It's very simple: the multiplicative subset $S$ contains $0$, since it's the ideal $2\mathbf Z/6\mathbf Z$. In this case, any $(r,s)\sim~(0,1)$, so $S^{-1}R$ is the null ring.
That's why, in the construction of rings of fractions, one usually requires the multiplicative subset does not contain $0$.
But you may localise at the prime ideal $2\mathbf Z/6\mathbf Z$. In this case, you can check $(\mathbf Z/6\mathbf Z)_{2\mathbf Z/6\mathbf Z}$ is isomorphic with the field $\mathbf Z/2\mathbf Z$, and the canonical map \begin{align} \mathbf Z/6\mathbf Z&\longrightarrow (\mathbf Z/6\mathbf Z)_{2\mathbf Z/6\mathbf Z}\\ \bar x&\longmapsto\frac{\bar x}{\bar 1} \end{align} is not injective: $\; \bar 0,\,\bar 2,\,\bar4$ map to $\dfrac{\bar 0}{\bar 1}$ and $\;\bar1,\,\bar 3,\,\bar 5$ map to $\;\dfrac{\bar1}{\bar1}$.