I need to prove two things
(a) prove or disprove that $S$ is a subring of $R$
(b) prove or disprove that $S$ is a group with respect to multiplication in $R$
For (a) (1) it is clearly that $S$ is non empty since it contains the identity (The unity)
(2) let $a,b \cap S$ then $ab \in S$ because for $a$ we know there is an element $x$ such that $ax = e$ and for $b$ we know what there exists an element $y$ such that $by = e$ and so $ab(yx) = a(by)x = aex = ax = e$ and so $ab \in S$
$a - b \in S$ i am stuck here ! any suggestions
for (b) i think we already proved that it is closed under multiplication and have the identity and it is trivial to prove that all elements have inverses and are also associative , so yes it is a group under multiplication
For the case (a), you already mentioned that $S$ is non-empty as $1 \in S$. But $1 - 1 = 0$ does not belong to $S$, unless $1 = 0,$ i.e. unless $R$ is the trivial ring. So if $R$ is non-trivial then $S$ can not be a subring.