I think the question loses some of its appeal once we note that the preorder $T\sqsubseteq T'$ defined by the OP coincides with the preorder defined by the inclusion $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ when restricted to theories which are strong enough to prove say $\exists!x\forall y(y\not\in x)$ (a formula suggested by Francois' comment). So if we restrict our analysis to theories which are closed under deduction ($\{\phi:T\vdash\phi\}=\mathtt{Th}(T)=T$), this preorder is simply the usual inclusion and if we focus on consistent theories (as Noah suggests) we are simply asking if a consistent theory can be extended to a maximal consistent theory (yes it can and it will be complete hence not recursively axiomatizable). So in the light of this, I think Noah's answer could be made simpler (This should really be a comment of mine, but my rep doesn't allow me to use that option). To see that $T\sqsubseteq T'$ is equivalent to  $\mathtt{Th}(T)\subseteq\mathtt{Th}(T')$ (for strong enough theories), the hardest part is to focus on $\Rightarrow$. So assuming that $T\sqsubseteq T'$ and $\phi$ is a sentence such that $T\vdash \phi$, hence $T\vdash\exists !x\phi\land\forall y(y\not\in x)$ from which we obtain $T'\vdash\exists !x\phi\land\forall y(y\not\in x)$ and finally $T'\vdash\phi$.