$\text{S}$ is reflexive: $(a,b)\text{S}(a,b)$ because $a-b=a-b.$
$\text{S}$ is symmetric: Suppose $(a,b)\text{S}(c,d).$ Then $a-b=c-d.$ But then also $c-d=a-b,$ so $(c,d)S(a,b).$
$\text{S}$ is transitive: Suppose $(a,b)S(c,d)$ and $(c,d)S(e,f).$ Then $a-b=c-d$ and $c-d=e-f.$ But then by transitivity of equality $a-b=e-f.$ So $(a,b)\text{R}(e,f).$
$\text{S}$ is reflexive, symmetric and transitive, so it is an equivalence relation.
For an equivalence relation $\text{R}$ on a set $\text{A}$, the set of the elements of $\text{A}$ that are related to an element, say a, of $\text{A}$ is called the equivalence class of element $a$ and it is denoted by $[a].$
Each equivalence class is a subset of the base set $\text{Z} \times \text{Z}$ on which the relation has been defined.
Each equivalence class has ordered pairs with the same result of subtracting the second element from the first and there is one equivalence class for each possible difference: $\dots, [(-2,0)], [(-1,0)], [(0,0)], [(1,0)], [(2,0)], \dots $ are all distinct equivalence classes.
NOTE that in any equivalence relation $\text{R}$ on set $\text{A}$, equivalence class of an element $x \in \text{A}$ with respect to $\text{R}$ i.e. $[x]$ is same as equivalence class of element $y \in \text{A}$ with respect to $\text{R}$ i.e. $[y]$ if and only if $x\text{R}y.$
So, for this question, say for Option C, we only need to check whether $(a,b)\text{S}(-b,-a),$ which is true as $a-b = -b -(-a)$
NOTE that in any equivalence relation $\text{R}$ on set $\text{A},$ equivalence class of an element $x\ in \text{A}$ with respect to $\text{R}$ i.e. $[x]$ is disjoint from the equivalence class of element $y \in \text{A}$ with respect to $\text{R}$ i.e. $[y]$ if and only if $x$ is Not related to $y.$
