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), 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 $M$ (we even have the strengthening described in this question) 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. (Added September 17, 2022) No, that doesn't work because, if $j: P \to M$ is a weakly compact embedding with critical point $\kappa$, $M$ may contain functions that are not in $P$. However, the argument does work for weakly Ramsey cardinals.
(Added September 17, 2022) A weakly superstrong Woodin cardinal is weakly Shelah. By theorem 5 of the paper that defined weakly superstrong cardinals, a cardinal $\kappa$ is weakly superstrong if and only if for any $A \subseteq V_\kappa$, there are $\lambda$ and $A^* \subseteq V_\lambda$ such that $\langle V_\kappa, \in, A \rangle \prec \langle V_\lambda, \in, A^* \rangle$. If $\kappa$ is Woodin and weakly superstrong, then, for any $f : \kappa \to \kappa$, there is a cardinal $\nu \lt \kappa$ that is $\lt \kappa$-strong for $f$, and an elementary extension $\langle V_\kappa, \in, A \rangle \prec \langle V_\lambda, \in, f^* \rangle$. By elementarity, $\nu$ is $\lt \lambda$-strong for $f^*$. Since this works for any $f : \kappa \to \kappa$, this shows that $\kappa$ is weakly Shelah and also satisfies the strengthening mentioned above.
(Added October 31, 2023) I have proved in another MothOverflow answer that $\kappa$ is weakly Shelah if and only if it is Woodin and there is a $\theta$ such that $V_\kappa$ is a $\Sigma_3$-elementary submodel of $V_\theta$ and that $\kappa$ satisfies my strengthened definition if and only if if is Woodin and weakly superstrong.