...as long as you also have boundary conditions specified at $x=\pm1$

You can treat this as the generator for a markov process in $(x,t)$ (introducing a separate random variable for the time parameter), from which you can see that there will be a unique solution with the piecewise initial conditions.

@yemon: the signed measure also vanishes on the constant functions. So, take its positive and negative components and scale them to get probability measures.

@Davide Giraudo: Interesting. Applying the Hahn-Banach theorem, that answers the question. $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac1n=\infty$.

@JosephO'Rourke: All but countably many positive reals are generic, as two distinct analytic functions can only intersect at countably many points.

I find this very surprising (mainly because I thought I had proven the opposite while thinking about this question). I'm going to have to have a look at Kechris's book in more detail.