Is $S=\{\overline{0},\overline{5},\overline{10}\}\subset\mathbb{Z}_{15}$ isomorphic to$\mathbb{Z}_{3}$ as rings? My book asked me to prove that it is in fact. I generated the tables for addition
\begin{array}{c|ccc}
+ & 0 & 5 & 10 \\
\hline 0 & 0 & 5 & 10 \\
5 & 5 & 10 & 0 \\
10 & 10 & 0 & 5
\end{array} \begin{array}{c|ccc}
+ & 0 & 1 & 2 \\
\hline 0 & 0 & 1 & 2 \\
1 & 1 & 2 & 0 \\
2 & 2 & 0 & 1
\end{array}
and for the product
\begin{array}{c|ccc}
\cdot & 0 & 5 & 10 \\
\hline 0 & 0 & 0 & 0 \\
5 & 0 & 10 & 5 \\
10 & 0 & 5 & 10
\end{array}
\begin{array}{l|lll}
. & 0 & 1 & 2 \\
\hline 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 2 \\
2 & 0 & 2 & 1
\end{array}
If I choose 0 $\mapsto 0$, $10 \mapsto 1$, because 10 and 1 are units of those rings, respectively and $5 \mapsto 2$, there is no problem to prove that $f(xy)=f(x)f(y)$, but
$$f(5+5)=f(10)=1 \neq 2=1+1=f(5)+f(5)$$
Did I do anything wrong?

You are almost right.
$f(x+y)=f(x)+f(y)$ also holds as
$$f(5+5)=f(10)=1  =2+2=f(5)+f(5)$$
The remaining argument to prove the isomorphism is to note that $f$ has inverse.
