I have solved this problem before but only just found it was from the JuniorTST from Romania 01 just there. Anyway since $A$ is non empty let $a$ be an arbitrary element. Then $a\in A\implies a+0\in A\implies (a)(0)=0\in A$. Then for all $x$ we have $x+(-x)=0\in A$ so their product, $-x^2$ belongs to $A$. This covers all negative reals. Then necessarily for $x>0$ we have $-2\sqrt{x}\in A$ i.e. $(-\sqrt{x})+(-\sqrt{x})\in A$ so their product, $x$, covers all positive reals. So all positive reals are in A, all negative reals are in A, and also $0$ is in A so $A=\mathbb{R}$.

A similar way to finish is to use the fact that $-x^2=-\frac{x^2}{2}-\frac{x^2}{2}\in A{\implies} \frac{x^4}{4}\in A$, which clearly covers all of the positive reals.