In some sense the empty set ($\emptyset$) and the global set of all sets ($G$) are the ends of the universe of mathematical objects. The world which $ZFC$ describes has an end from the bottom and is endless from the top. Even in a straight forward way one can find an equiconsistent theory (respect to $ZFC$) which its world is endless from the bottom and bounded from the top by the set of all sets. It is sufficient to consider the theory $ZFC^{-1}$ ($ZFC$ inverse) which is obtained from $ZFC$ by replacing each phrase $x\in y$ in the axioms of $ZFC$ by the phrase $\neg (x\in y)$. This operation for example transforms the axiom of empty set of $ZFC$ to an statement which asserts "the set of all sets exists".
$[\exists x \forall y~~\neg(y\in x)]\mapsto [\exists x \forall y~~\neg \neg(y\in x)] $
Even the axiom of extensionality remains unchanged because we have:
$[\forall x\forall y~~(x=y\longleftrightarrow \forall z~~(z\in x\longleftrightarrow z\in y))]\mapsto [\forall x\forall y~~(x=y\longleftrightarrow \forall z~~(\neg (z\in x)\longleftrightarrow \neg (z\in y)))]$
So the "set of all sets" is unique in this theory. Even the equiconsistency simply follows from the fact that for all set (or proper class) $M$ and for all binary relation $E$ on it we have:
$\langle~M~,~E~\rangle \models ZFC \Longleftrightarrow \langle~M~,~M\times M\setminus E~\rangle \models ZFC^{-1}$
So it is trivial that $ZFC^{-1}\models \neg (\exists x \forall y~~\neg(y\in x))$ in the same way which one can prove $ZFC\models \neg (\exists x \forall y~~y\in x)$ by the Russell's paradox. But the situation seems rather strange when one wants to find an equiconsistent theory with $ZFC$ which has end points in both up and down direction because the existence of two contradictory objects like $\emptyset$ and $G$ seems ontologically incompatible in a particular "$ZFC$-like" world. So the question is:
Question (1): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:
$(1)~Con(ZFC)\Longleftrightarrow Con(T)$
$(2)~T\models \exists !x~\forall y~~(y\in x)$
$(3)~T\models \exists !x~\forall y~~\neg (y\in x)$
Remark (1): Quine's new foundation axiomatic system ($NF$) is not an answer because its equiconsistency with $ZFC$ is still unknown.
Even one can define two dual sets from empty and global sets. The set which does not belong to any other set ($\emptyset^{\star}$) and the set which belongs to any set ($G^{\star}$).Now one can restate the question (1) as follows:
Question (2): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:
$(1)~Con(ZFC)\Longleftrightarrow Con(T)$
$(2)~T\models \exists !x~\forall y~~(x\in y)$
$(3)~T\models \exists !x~\forall y~~\neg (x\in y)$
Even it is interesting to have an equiconsistent theory which has no end points in both up and down directions.So:
Question (3): Is there an $\mathcal{L}=\lbrace \in\rbrace$-theory $T$ such that the following conditions hold:
$(1)~Con(ZFC)\Longleftrightarrow Con(T)$
$(2)~T\models \neg (\exists x~\forall y~~(y\in x))$
$(3)~T\models \neg (\exists x~\forall y~~\neg (y\in x))$
Best Answer
Here's an answer which is a bit better than the one I suggested in the comment, but unfortunately still quite unnatural.
Question 1
Define $U(x) := \forall y\;y\in x$. Then adjust the axioms of $\mathsf{ZF}$ (other than extensionality) so that they are "bounded over sets that aren't universal." That is, replace every universal quantifier $\forall x \;\ldots$ with $\forall x\;\neg U(x) \rightarrow \ldots$ and every existential quantifier $\exists x \; \ldots$ with $\exists x \; \neg U(x) \wedge \ldots$. Then keep the extensionality axiom the same and add the axiom $\exists x\;U(x)$. Call this theory $\mathsf{T}$. Note that we can easily produce models of $\mathsf{T}$ from models of $\mathsf{ZF}$ by adding one extra element for the universal set. We can produce models of $\mathsf{ZF}$ from models of $\mathsf{T}$ by removing the universal set.
Question 2
This follows from the answer to question 1 by replacing $x \in y$ by $y \in x$ in every axiom of $\mathsf{T}$.
Question 3
Define $Q(x) := \forall y\;y \in x \leftrightarrow y = x$. That is, $Q(x)$ says that $x$ is a Quine atom. Let $\mathsf{T}$ be the theory with the axiom $\exists ! x \;Q(x)$ and with the axioms of $\mathsf{ZF}$ adjusted as follows. Like in question 1, quantifiers should be bounded to non-Quine atoms. This time we also require that whenever $\mathsf{ZF}$ would assert the existence of a set, $\mathsf{T}$ asserts the existence of the same set but with a Quine atom added. So for example, empty set becomes $$ \exists x \; \neg Q(x) \wedge (\exists z \; Q(z) \wedge z \in x) \wedge (\forall y \; \neg Q(y) \rightarrow \neg y \in x) $$ Separation would become $$ \forall x\,\neg Q(x) \rightarrow (\exists y\, \neg Q(y) \wedge (\exists z\; Q(z) \wedge z \in y) \wedge \forall z \; (\neg Q(z) \rightarrow (z \in y \leftrightarrow (\,z \in x \wedge \phi(z)\,)))) $$ Similarly to question 1, we can convert between models of $\mathsf{ZF}$ and models of $\mathsf{T}$ by just adding or removing a single element (this time corresponding to the Quine atom).