Let's interpret your question as asking for the quantifier complexity of the assertion "$\kappa$ is a Woodin cardinal".

More specifically, one natural way to interpret your question, whether $\kappa$ is Woodin if and only if "there is $S$ such that..." is to interpret the ellipses as having only bounded quantifiers. This amounts to asking whether "$\kappa$ is Woodin" is a $\Sigma_1$ property of $\kappa$. For this, the answer is no, this level of complexity is much too simple, because $\Sigma_1$ properties are trivially preserved upwards to larger models, and we can destroy the Woodin-ness of $\kappa$ by forcing to collapse it, for example. So in this sense, the property of $\kappa$ being a Woodin cardinal cannot be witnessed by the existence of a single set $S$. (Of course, it is witnessed in one direction, as Andres explains in the comments, in that if $\kappa$ is Woodin, then it is Woodin in $L[S]$ for some set $S$, but this is not an equivalence unless one imposes requirements on $S$ that increase the complexity.)

Meanwhile, the property expressing that $\kappa$ is Woodin is $\Delta_2$. It is $\Sigma_2$, because as Andres points out in the comments, $\kappa$ is Woodin just in case there are extenders witnessing all the required $\lt\kappa$-strongness embeddings that arise in the definition of being Woodin, and so any $V_\alpha$ with $\alpha>\kappa$ can see that $\kappa$ is Woodin. Any such property is $\Sigma_2$. For the same reason, the property of not being Woodin is also visible inside any such $V_\alpha$, and so not being Woodin in also $\Sigma_2$. So the assertion "$\kappa$ is a Woodin cardinal" has complexity $\Delta_2$.