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][1] 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:

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

> 2. 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). Then the moments of $\mu-m^+$ and $\nu+m^-$ coincide on $S$, and we cannot have both equalities $\mu=\nu$ and $\mu-m^+=\nu+m^-$.

  [1]: http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem