If $|X|>2$, then there is a single solution: $\overline{\mu}(\{x\})=\mu(x)$ and $\overline{\mu}(S)=0$ for any other set $S$.
Indeed, let $a=\overline{\mu}(\emptyset)$. By the condition for $S=\{x\}$, $\overline{\mu}(\{x\}) =\mu (x)-a$. By the condition for $S=\{x,y\}$ with $x\neq y$, $\overline{\mu}(\{x,y\}) =a$.
Now if $|X|>2$, there are at least $1+|X|$ two-element subsets. Thus, $\sum_{|S|\leq 2} \overline{\mu}(S) \geq a+ \sum_{x\in X}\mu (x) =1+a$. This forces $a=0$.
Now it is easy to show that for all larger subsets $\overline{\mu}(S)=0$.