Let $R$ a commutative ring and $S\subset R$ a subset closed multiplicatively with $1\in S$. We say that $(r,s)\sim (r',s')$ if there is a $t\in s$ s.t. $t(r's-rs')=0$. The thing is I don't understand the presence of this $t$, and I don't find any example where it's important. I tried for example for $R=\mathbb Z/6\mathbb Z$ and $S=\{1, a\}$ where $a\in \{1,...,5\}$, then with $S=\{0,1,2,3\}$, but still, it doesn't work. What I thought was something like: $$\frac{2}{1}=\frac{6}{3}$$ but since $\frac{6}{3}=\frac{0}{1}$, we would have that $\frac{2}{1}=\frac{0}{1}$, but is there something around that ?
By the way, it's not precise in my course that $0\notin S$ (that's why I included it), but may be it can't be in $S$, no ?
Take $\mathbb{Z}/6\mathbb{Z}$ and the set $S=\{1,2,4\}$, which is multiplicatively closed. Then $$\frac{5}{2}=\frac{1}{1}$$ since $2\cdot(5-2)=2\cdot 3=0$. And $3\neq 0$ in $\mathbb{Z}/6\mathbb{Z}$.
So $(5,2)\sim(1,1)$ but $5\cdot 1- 1\cdot 2\neq 0$.
If $0\in S$ then $S^{-1}R=0$ that's why some prefer to impose $0\not\in S$.