This question was answered in the comments:

As Gabe Goldberg pointed out, Trevor Wilson's argument that $\Sigma_3$-reflecting Woodin cardinals are weakly Shelah fails because the reflection argument would have to use the function $f$ as a parameter, and $\kappa$ being $\Sigma_3$-reflecting is not enough for that.

Additionally, as Sean Cox pointed out, we can't be sure that $V_\kappa \prec V_{j(f)(\kappa)}$ since it is possible that there is some $\beta \lt \kappa$ such that $j(f)(\beta) \gt \kappa$. I assumed that $j(f) \upharpoonleft \kappa = f$ but I don't know how to prove that from the definition.