I don't have a complete proof yet, but I have a plausible conjecture. Let $\mu$ be a probability measure in the plane, define the potential $$u(z)=\int\log|1-z/t|d\mu(t).$$ Then I conjecture that $\mu\in M$ iff $u$ satisfies $u(z)\leq u(|z|)$ for all $z$.
It is evident that this condition is necessary. It seems that it is strictly stronger than the Obrechkoff condition. I don't think that this condition can be restated as a simple property of $\mu$ itself.
To prove the sufficiency, I am first going to restrict to a dense subclass of $\mu$ with convenient properties (it is clear that it is enough to prove sufficiency for a dense subclass). The convenient properties I have in mind is that $\mu$ does not charge some small angular sector $|\arg z|<\epsilon$ and that it behaves nicely near $0$ and $\infty$, say has some small atom at $-\epsilon$ and nothing else in the disc $|z|<100\epsilon$, and similarly at infinity. In addition, I want to require that $u(|z|)>u(z)$ for all $z$ except on the positive ray.
Then I am going to discretize the measure to obtain a polynomial, whose $(1/n)\log|P_n|$ approximates $u$ nicely near the positive ray, and apply the saddle point method to the integral $$\int_{|z|=r}\frac{P_n(z)}{z^k}\frac{dz}{z},$$ with $n\to\infty$, using the nice behavior near the positive ray, and obtain an asymptotic for the coefficients which will show that they are positive.
The difficulty is that the asymptotics must be uniform in $k$, but I hope to achieve this by the arrangement near $0$ and $\infty$ described above.
In fact, there is an (unpublished and unproved) conjecture of Alan Sokal that if a polynomial satisfies $|P(z)|<P(|z|)$ then some sufficiently high power has positive coefficients. This of course would imply sufficiency of my condition.
ADDED on July 19. The above outline is correct; we are writing a proof which will soon be posted on arxiv.
ADDED on August 23. Here is the precise statement. A probability measure $\mu$ is a limit measure if and only if it is symmetric with respect to complex conjugation, and $u(z)\leq u(|z|)$ where $$u(z)=\int_{|z|\leq 1}\log|z-\zeta|d\mu(\zeta)+\int_{|z|>1}\log|1-z/\zeta|d\mu(\zeta).$$ (The potential I wrote earlier may be divergent for some probability measures, so it has to be modified a little bit).
