If $G$ is a finite group and all non-identity elements of $G$ are order 2 then the product of these elements is the identity.

As you note, the group is abelian. Since it has no element of order $p$ for any odd prime $p$, the order of the group is divisible only by the prime $2$, so the order is a power of $2$.

The group then has a subgroup $H$ of index $2$, with a coset $aH$. To each element $h$ of $H$ there corresponds an element $ah$ of $aH$. The product of these two elements is $ah^2$, which is $a$. So the product of all the elements is $a^r$, where $r$, half the order of the group, is even, and we're done.

Write $G=\displaystyle\bigcup_{g\in G }gH $.

For $g_1,g_2\in G $, either $g_1H=g_2H $ or $g_1H\cap g_2H=\emptyset$. Let $G'$ be a set of all $g$'s with pairwise disjoint cosets. Then we can write $G=\displaystyle\bigcup_{g\in G' }gH.$

Now the product of all elements can be written as $$\prod_{g\in G'}g\cdot ga\cdot gb\cdot gab =\prod_{g\in G'}g^4\cdot (ab)^2=1(\text{since $G$ is Abelian and every element is of order $2$}). $$