SAUDI ARABIAN MATHEMATICAL COMPETITIONS

If $a=0\in S$ or $a=1\in S$ then for any $b\in S$ we have $ab\in S$ which is obvious. So we can suppose that $a,b\notin \{0;1\}$. From $1\in S$, $a\in S$ we have $1-a\in S\to (1-a)-1=-a\in S$, then $$b-(-a)=a+b\in S.$$ We also have $a-1\in S$ so $\frac{1}{a},\frac{1}{a-1}\in S$ then $$\frac{1}{a-1}-\frac{1}{a}=\frac{1}{a(a-1)}\in S\to a(a-1)\in S.$$ Since $a\in S$, $a(a-1)\in S$ then $a^2=a(a-1)+a\in S$. Similarly, we also have $b^2\in S$. Thus $a,b\in S\to a+b\in S\to (a+b{)}^2\in S$. Finally, $(a+b{)}^2-a^2=2ab+b^2\in S$ and $\left(2ab+b^2\right)-b^2=2ab\in S$ which implies that $$\frac{1}{2ab}+\frac{1}{2ab}=\frac{1}{ab}\in S\to ab\in S.$$