Consider non-commuting variables $x_1,x_2,\dots,x_n$. For each subset $S\subseteq \{1,2,\dots, n\}$, define the following monomial: $$(-1)^lx_{s_1}x_{s_2}\dots x_{s_k}x_{t_l}x_{t_{l-1}}\dots x_{t_1},$$ where $S=\{s_1<s_2<\cdots<s_t\}$ and $\{1,2,\dots,n\}\setminus S=\{t_1<t_2<\cdots<t_l\}$.
Informally, the monomial is made by multiplying all the $x_i$'s for $i\in S$ in increasing order of indices and then the $x_i$'s for $i\not\in S$ in decreasing order of indices and multiplying the monomial by a sign depending upon the size of the set $\{1,2,\dots, n\}\setminus S$.
I claim that the sum of such monomials indexed by all possible subsets $S$ of the set $\{1,2,\dots, n\}$ is zero.
For example, when $n=3$, we have the following: $$S=\phi: -x_3x_2x_1$$ $$S=\{1\}: x_1x_3x_2$$ $$S=\{2\}: x_2x_3x_1$$ $$S=\{3\}: x_3x_2x_1$$ $$S=\{1,2\}:-x_1x_2x_3$$ $$S=\{1,3\}:-x_1x_3x_2$$ $$S=\{2,3\}:-x_2x_3x_1$$ $$S=\{1,2,3\}:x_1x_2x_3$$ Then, we can see that the sum is indeed zero.
I know two ways of proving this. One is a proof by induction on $n$ which is quite short and straightforward, but isn't really that illuminating and interesting. Another is a high-tech proof which is by using properties of Hopf algebras: One can think of these non-commuting variables to be generators of a free algebra, and then, the above sum of monomials reduces to one of the defining axioms of Hopf algebras relating the antipode to the multiplication and the comultiplication.
But really, this seems like a combinatorial statement and it would be nice to have some kind of bijection/counting argument to show why this sum is zero.
I'll use the same notations as you. For each $S=\{s_1,\dots,s_t\}$, you can define $S'$ by either $S':=S\cup \{n\}$ if $n\not\in S$ or $S':= S\setminus \{n\}$ if $n\in S$.
Then $S\mapsto S'$ is a bijection on $\mathcal{P}(\{1,\dots,n\})$ (indeed, $S''=S$), and the monomials corresponding to $S$ and $S'$ cancel each other (they have opposite signs). Therefore the whole sum is zero.