Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 2 / a / i / Solution

The localization is the zero ring exactly when $1/1=0/1$, which means that some $u\in S$ is zero. Thus $0\in S$. This holds exactly when $S$ contains a nilpotent element: if $s^n=0$ and $s\in S$, multiplicative closure gives $0\in S$, while zero itself is nilpotent.