$S^{-1}R=0$ iff $0$ belongs to $S$. Let $R$ be a commutative ring with $1$ as identity and let $S$ be a multiplicative closed subset of $R$ containing $1$. Show that $S^{-1}R=0 \text{ iff } 0$ belongs to $S$.

My approach: Now $S^{-1}R=0$ implies if the $1$ in this ring is $zero$ i.e. $(1,1)~(0,1)$. This occurs if $x.1=0$ for some $x$ in $S$. That is if $0$ belongs to $S$. Now conversely let $0$ belongs to $S$. Then help me to procced in this direction.
Thanks for all your help in advance.
Recall that $(x,s)\simeq (x',s')$ iff $s'x=sx'$. The other way $(1,1)\simeq (0,0)$ since $1.0=0.1$
Recall that $(x,s)\simeq (x',s')$ iff $s'x=sx'$. The other way $(1,1)\simeq (0,0)$ since $1.0=0.1$