Let $A$ be a finite alphabet, and let $w \in (A \cup A^{-1})^\ast$ be a freely reduced word over the alphabet $A$ and formal inverse symbols $A^{-1}$. Suppose $w$ is non-empty. Can there ever be non-empty prefixes $p_i \in (A \cup A^{-1})^\ast$ of the word $w$ such that $p_1 p_2 \cdots p_n = 1$ in the free group on $A$?

For example, if $w = abca^{-1}$ then it is easy to see that no product of elements from $\{ a, ab, abc, abca^{-1}\}$ ever equals $1$. Some "non-trivial" behaviour can happen: for example, if $w = aba^{-1}b^{-1}a^{-1}$ then we find the product $$aba^{-1}b^{-1}a^{-1} \cdot aba^{-1} \cdot ab = aba^{-1}b,$$ i.e. the word $aba^{-1}b$ is a product of three prefixes, but is not graphically a product of prefixes, of $w$. I don't see how this kind of cancellation could ever be enough to completely cancel such a product, however.

An equivalent formulation of the question is: let $X_w$ be the subsemigroup of a free group $F_A$ generated by the set of non-empty prefixes of a word $w \in F_A$. Can we ever have $1 \in X_w$?