Suppose that $R$ is a commutative ring and $S\subset R$ is multiplicatively closed subset, i.e. $1\in S$ and if $a,b\in S$ then $ab\in S$. Consider the natural mapping $\phi:R\to S^{-1}R$ defined by $\phi(r)=\frac{r}{1}$.
If $S$ does not contain zero-divisors then $\phi$ is injective.
I am trying to prove it directly but have some troubles.
My approach: Let $r\in \operatorname{Ker} \phi$ then $\phi(r)=\frac{0}{1}$. It means that $\frac{r}{1}=\frac{0}{1}$ then by definition it means that $\exists u\in S$ such that $u(r\cdot 1-0\cdot 1)=0$, i.e. $ur=0$. How to conclude that $r=0$?
If $u\neq 0$ then $ur=0$ leads to $r=0$.
What if $u=0$?
Would be very grateful for help!
If $R$ is a nonzero ring, then $0$ is a zero divisor in $R$, so $0\not\in S$ since by assumption $S$ contains no zero divisors. So since $u\in S$, we cannot have $u=0$. (If, as you suggested in a comment, your definition of "zero divisor" does not include $0$ as a zero divisor, then the statement you are trying to prove is not true and you need to modify it to also require $0\not\in S$.)
If $R$ is the zero ring, then the conclusion that $r=0$ is trivial since $0$ is the only element of $R$.