True or False: If the product of n elements of a group is the identity element, it remains so no matter in what order the terms are multiplied.

Since this is a true or false question, it is not that the question is phrased incorrectly, but rather that the answer is that it is false.

Your claim that it can only hold if the group is abelian is not true for all such $a, b, c$, which we can see in any group by $a=b=c=e$ and other less trivial examples. What you need to do to show it is false is to pick a specific nonabelian group and find three specific elements where it doesn't work.

Edit: I just noticed that this was not an exercise, but part of the solution manual. In $S_3$, the elements $(12),(23),(321)$ show the original claim is invalid. In fact, if $abc=e$ and $bc\neq cb$, then it is never true that $acb=e$ because $bc$ is the unique inverse of $a$. By similar reasoning, if the claim holds for $a, b, c$, then all three elements commute pairwise. You can show that the group is abelian if this holds true for all triples such that $abc=e$.

What I was expecting the statement to say was:

If the product of $n$ elements in the group is the identity, it remains so under any cyclic permutation.

Perhaps this is what “order” is meant in the solution? In any case, this might be a reasonable statement to prove for yourself.

Counter examples could be found in $2\times 2$ invertible matrices.

For example $A= \begin{bmatrix}1&2\\2&1\end{bmatrix}$,$B= \begin{bmatrix}3&2\\1&5\end{bmatrix}$ $C=(AB)^{-1}$ then obviously $$ABC=I$$ while $$CBA\ne I$$