This would be right except that it is not necessarily true that $S_M=0$ on the event $X_1 > 0$. If $p=2/3$, for example, then $X$ takes values $-1$ and $1/2$, So $S_M$ can take the value $-1/2$.

I think Ryan's solution works and gives the $2\pi+h+\epsilon$ that drvitek originally claimed.

I put in a simpler "example" than the one I initially found but it was too simple. The modified one should work.

Yes I see now. Sorry for being confused! Actually this is already an interesting question in the discrete case. Does there exist a sequence $(X_n)_{n \in \mathbb{N}}$ of random variables with $X_{j+1} + \ldots +X_{j+n}$ having Binomial$(n,1/2)$ distribution for all $j$ and $n$, which is not simply a sequence of independent binomial random variables?

I've elaborated on my answer. If I'm misunderstanding something please let me know.

With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)-X(p)$ is a non-negative integer. Since $X$ is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued, so it is a point process.

For generalizations it's useful to be able to take either perspective: (A) and (B) as a single "niceness" condition; or, separate them and try to relax them separately.

For Lévy's theorem you only need it to be a local martingale.