Does absolutely flat commutative ring imply all ideals are idempotent?

I think they are saying this. Given an ideal $I$, tensor the inclusion $$ 0 \to I \to R $$ with $R/I$. We get (flatness of $R/I$) $$ 0 \to I/I^2 \to R/I. $$ Thus $I/I^2 = 0$, so $I = I^2$.

This holds for noncommutative rings as well. If you are aware that "absolutely flat" is sometimes called "von Neumann regular", then you know that every finitely generated ideal of $R$ is generated by an idempotent.

If $I$ were an ideal with $I^2\subsetneq I$, pick $x\in I\setminus I^2$. Then $(xR)^2=(eR)^2=eR=xR$. But this means that $x\in (xR)^2\subseteq I^2$, a contradiction. So, $I^2=I$.