Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$. 
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.

If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab \mid b^2$. Now only remains to show that $ab \mid -2ab$ which is clearly true.
Is this valid? I'm not sure about the implication that $ab$ would divide all the terms in $a^2+2ab+b^2$.

Not quite: in general, for $a,b,c\in\mathbb{Z}$, $a|b+c$ does not imply $a|b$ and $a|c$.
Instead, simply suppose $ab|a^2+2ab+b^2$. That is, $abk=a^2+2ab+b^2$ for some $k\in\mathbb{Z}$. The rest sort of follows like the end of your idea, if you subtract $4ab$ from both sides of the equation, then $abk-4ab=a^2-2ab+b^2$, so clearly $ab|(a-b)^2$.
