I think any measurable Woodin cardinal is a limit of weakly Shelah cardinals.

To see this, note that, if $\kappa$ is a Woodin cardinal, for any $f : \kappa \to \kappa$, $\kappa$ is a limit of cardinals $\nu$ that are $\lt \kappa$-strong for $f$ (that's probably a standard result but I know it from theorem 9.9. of ["Double helix in large large cardinals and iteration of elementary embeddings" by Kentaro Sato](https://doi.org/10.1016/j.apal.2007.02.003)), which means that for any $\gamma \lt \kappa$, there is an elementary embedding $\bar i : V \to \bar N$ such that $V_{f(\gamma) +1} \subset \bar N$ and $\bar i(f) \upharpoonleft (\gamma +1) = f \upharpoonleft (\gamma +1)$.

If $\kappa$ is additionally measurable, there is an elementary embedding $j : V \to M$ with critical point $\kappa$. By elementarity, there's an elementary embedding $i : M \to N$ with critical point $\nu$ such that $V_{f(\kappa) +1}^M \subset N$ and $i(f) \upharpoonleft (\gamma +1) = j (f) \upharpoonleft (\gamma +1)$. Since this works for any $f : \kappa \to \kappa$, this shows that $\kappa$ is weakly Shelah in $M$ and thus a limit of weakly Shelah cardinals.

By a similar argument using the elementary embedding characterization of weakly compact cardinals, we can probably prove that any weakly compact Woodin cardinal is a limit of weakly Shelah cardinals.