This question is inspired by this question.
Given a finite group $G$, is there an ordering $G=\{a_1, \dots, a_n\}$ of its elements such that the product of all group elements in that specified order equals the first element, i.e $a_1\cdot\dots\cdot a_n=a_1$. Equivalently, can we multiply all but one element of the group in such a way to obtain the unit $1$.
Trivial cases (probably not very helpful):
$G$ is abelian (just define $a_1$ to be the product of all elements).
$G$ has no element of order $2$ ($a_1=1$ and pair the elements with their inverses in the list).
$G$ has a unique element of order $2$ (set $a_1$ to be that element and pair the others with their inverses).
Slightly less trivial cases (still probably not that useful)
$G=S_3$ (symmetric group) if $a=(12), b=(23)$ are the standard generators, then $(aba)=(aba)(a)(ba)(b)(ab)1$
$G$ with $|G|>6$ has exactly two or three elements of order $2$: Consider the conjugation action of $G$ on the set $X$ of elements of order $2$. If this action is non-trivial, then for at least one $x\in X$ the centralizer of $x$ has at most $\frac{|G|}{2}$ elements, hence we find a $g\in G\setminus X$ with $y:= gxg^{-1}\neq x$; then the product $ygxg^{-1}$ followed by all the elements of order $\geq3$ paired with their inverses is trivial. If the action is trivial, then $X$ lies in the center and is thus an abelian subgroup (because the product of commuting elements of order $2$ has order $\leq 2$); hence we can first multiply all elements of $X$, followed by all other elements paired with their inverses.