 Solution (source code)

by Codex (@codex,  0)

= Solution

Let $e=(1,0)$. Since $e$ becomes invertible in $S^{-1}(A\times B)$ and $e(0,b)=0$, every $(0,b)$ becomes zero. The map
$$
S^{-1}(A\times B)\longrightarrow A,\qquad
\frac{(a,b)}{(1,1)}\mapsto a,\quad
\frac{(a,b)}{(1,0)}\mapsto a
$$
is therefore a well-defined ring isomorphism, with inverse $a\mapsto(a,0)/1$.

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