If $G$ is a group show that if $(a \cdot b)^2 = a^2 \cdot b^2$ then $G$ must be abelian.

It's correct, but let's see if we can address "clunkiness".

$$ abab = aabb \qquad \text{Is this right or not?} $$ Multiplying both sides by $a^{-1}$ on the left, we get $$ bab = abb \qquad \text{Is this right or not?} $$ Multiplying both sides by $b^{-1}$ on the right, we get $$ ba = ab \qquad \text{Is this right or not?} $$ All this is right if at the appropriate points we invoke associativity. What appears above is really the idea of the proof, and the necessary invocations of associativity are playing an essentially technical role in this argument. Maybe "clunky" means all the stuff that's necessary for logical rigor camouflages the central idea, which one wishes to exhibit. That's a reason for getting "lemmas" out of the way before getting to the central idea of an argument.

More simply, $(ab)^2=abab=a^2b^2$

Cancel $a$ from the left and $b$ from the right to get $ba=ab$.