As $t$ is integral over $A$, there are $a_{1},\ldots,a_{r-1}\in A$ such that

$$t^{r}=a_{1}t^{r-1}+\cdots+a_{r-1}t+a_{r}.$$

We choose $s\in S$ such that $st^{i}\in A$ for $i=0,\ldots,r-1$. By induction on $n$, we prove that $st^{n}\in A$ for all $n\geq 0$. Note that

$$t^{n}=a_{1}t^{n-1}+\cdots+a_{r-1}t^{n-r+1}+a_{r}t^{n-r}.$$

Thus, if $st^{i}\in A$ for $i=0,\ldots,n-1$, then $st^{n}\in A$ because

$$st^{n}=a_{1}(st^{n-1})+\cdots+a_{r-1}(st^{n-r+1})+a_{r}(st^{n-r}).$$

∎