I'd just like to know where to start with this problem, where $R$ and $S$ are equivalence relations on non-empty set $A$, and suppose that $A-R = A-S$. I am looking to show that equivalence relations $R = S$.
If I say, let an example pair be an element of $A - R$. But with the exclusion of the set, how would you then make the steps to say that $R$ = $S$? Would it involve arbitrating something being equal to what's not removed?
As it was pointed out in the comments, equivalence relations are subsets of the cartesian product of the set, since they are formed via pairing elements with another elements. Then your statement has to be proven from $A\times A\setminus R=A\times A\setminus S$.
The statement is true, but not because $R$ and $S$ are equivalence relations, it is true because you're taking the set difference from the total set. What I mean is: If $X$ is a set and $A,B\subset X$ are subsets, then $A=B\Leftrightarrow X\setminus A=X\setminus B$. Since $X\setminus A$ is the complement of $A$ in $X$, let's denote it by $A^\mathsf c$ (same for $B$), this is the same as saying that $(A^\mathsf c)^\mathsf c=A$.
The $\Rightarrow$ part is obvious. Let's see the $\Leftarrow$ part: $x\in A\Leftrightarrow x\notin X\setminus A\Leftrightarrow^\text{by hypothesis} x\notin X\setminus B\Leftrightarrow x\in B$, so $A=B$.