No. $bb=aa$ is the same thing as $b^2=a^2$, which is the same thing as $e=e$ by the assumption that $a^2=b^2=e$. The statement that $e=e$ certainly is true, but does not imply that $a=b$ :)
For example, in $\mathbb{R}^\times$ under multiplication, $1^2=(-1)^2=1$, but $1\neq -1$.
Another example would be the group $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}=\{(0,0),(1,0),(0,1),(1,1)\}$ under addition: we have that $(1,0)+(1,0)=(0,0)=(0,1)+(0,1),$ but $(1,0)\neq(0,1)$.
Another example would be the symmetric group $S_3$, consisting of permutations of three objects, named $1$, $2$, and $3$. Consider the permutation $\sigma=(12)$ that switches $1$ and $2$, and does nothing to $3$, and the permutation $\tau=(13)$ that switches $1$ and $3$, and does nothing to $2$. Then $\sigma^2=\tau^2=$ the identity (the "do-nothing" permutation), because if you switch two objects and switch them again they end up where they started, but $\sigma\neq\tau$.