We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.

Recall Müntz-Szász theorem, which states (in particular) the following:

**Theorem:** If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.

the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.

If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.

If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.

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