the odd order case is easy, as the identity is the only self-inverse element. Just pair everything up with its inverse so they all cancel out, and let $b_0$ be the identity.

What is the point of changing the indexing from $a_i$ to $b_j$? What is the relation between $i$ and $j$ here?

@Pedro: the question amounts to asking if there is a $\sigma\in S_n$ such that $a_{\sigma(1)}\cdots a_{\sigma(n)} = a_{\sigma(1)}$.

@ArturoMagidin I suspected something like this...

Sorry, I don't get it, it is true that for odd order group, if $x*x=e$, then $x=e$, but how do you pair them up? @ZoeAllen

e.g. just write $aa^{-1}bb^{-1}cc^{-1} ... \cdot 1 =1$

wait, I mean $b_0, b_1, ..., b_n$ all have to be distinct. If you let $b_0=e$, then how do you assign $b_1$? Does $b_1=a_1*a_1^{-1}$? @ZoeAllen

I should have put $1$ aka $e$ at the start, to match your formulation.

$b_0=e$, $b_2=b_1^{-1}$, $b_4=b_3^{-1}, ..., b_n=b_{n-1}^{-1}$, ($n$ is even) I know all odd index elements are distinct, and all even elements are distinct, but how do you guarantee the odd index and even index elements are distinct? For example, maybe $b_4=b_1$? @ArturoMagidin

You pick them distinct. How do you know that $b_3$ is not $b_1$ or $b_2$? Because you picked it so that it wasn't any of $b_1$ or $b_2$. Then $b_4$ cannot be $b_1$ or $b_2$ either, because it is $b_3^{-1}$. Etc.

I've done some playing around and I think $S_3$ might be a counterexample.

@ZoeAllen I'm not sure it is. Sage says $$(2,3) \cdot e \cdot (1,3,2) \cdot (1,3) \cdot (1,2,3) \cdot (1,2) = (2,3).$$ Iirc sage uses right actions by default, which is worth keeping in mind if checking this by hand.

Suppose we are done for $b_1, b_2$, where $b_2=b_1^{-1}$, now we pick $b_3$ such that $b_3\neq b_1$ and $b_3\neq b_2$, but when I seek for $b_3^{-1}$, it might equal $b_1$. If I say okay, since $b_3^{-1}=b_1$, then I pick something else as $b_3$, then how do I know I always have something to pick when there is only a few elements left in the unpicked bank? @ArturoMagidin

No, it cannot happen, If $b_3^{-1} = b_1$, then $b_3=b_1^{-1}=b_2$, and you picked $b_3\neq b_2$. Do try to think it through a bit, won't you? If the inverse of $b_{2k+1}$ is among the previous pairs, then so is $b_{2k+1}$ itself.

@HallaSurvivor I'm not sure that works. I've gotten myself in a bit of a muddle, so I'm not sure, but isn't $(13)(123)(12)$ equal to the identity?

@ZoeAllen If you compose right to left, yes. If you compose left to right, then $1\mapsto 3\mapsto 1\mapsto 2$. He is composing left-to-right.

@ZoeAllen To get the equation with composition right-to-left, just replace each element with its inverse: $(2,3)e(1,2,3)(1,3)(1,3,2)(1,2) = (2,3)$.

Sorry for my silly question, I immediately realized this contradiction after I left home... Thank you!@ArturoMagidin

This is probably overkill, but it shows that unless the Sylow $2$ subgroup of $G$ is cyclic, there is always one such product equal to $e$, and so we can take $b_0=e$ and get an affirmative answer.