If, for every atom $a$, $a \subseteq x$ implies $a \subseteq y$, then $x \subseteq y$. Assume $x \nsubseteq y$, then by Supplementation, there is a $z \subseteq x$ such that $\neg z \ O \ y$. By Atomicity, there is an atom $a \subseteq z \subseteq x$, and $a \nsubseteq y$ because $\neg zOy$, contradicting the assumption that if $a \subseteq x$ then $a \subseteq y$.

This implies that the usual definition $a \subseteq b \iff \forall x (x \in a \Rightarrow x \in b)$ and your definition of $\in$ from $\subseteq$ are inverses of each other, so it does not matter whether the language is $\in$ or ($\subseteq$, $\{\}$); your mereological theory is synonymous with $\sf MMK$.

The axiom of Infinity doesn't actually say that there's an infinite set, just an infinite class (it doesn't say that $\{x\}$ exists), which is already a consequence of Composition. I didn't notice that until I had already written the entire answer. I've edited it in a few places to reflect which results actually need Infinity and which don't.

This theory doesn't seem to have a power set axiom, so it has a model in which all sets are countable. Specifically, the power set of the set of hereditarily countable sets, but without the empty set and with two Quine atoms, with the obvious $\in$ relation, is a model of MMK (with an infinite set). This implies that MMK is consistent, and because it is provably consistent in ZFC, it does not interpret MK, so it is not synonymous.

No extension of MMK with new axioms is synonymous with any subset of MK. A model of MK has no nontrivial definable automorphisms, but any model of MMK has a definable automorphism by swapping the two Quine atoms. If they were synonymous, then a model of MK would also be a model of MMK (with a different $\in$), and therefore have a nontrivial definable automorphism, which is impossible. (A similar argument might work to prove they are not bi-interpretable, but I'm not sure of the details). So the lack of power sets isn't the "main" problem; MMK is not synonymous with MK because we can't distinguish the two Quine atoms. (The reason this argument didn't work in the case of ZFGC was the global choice function $C$, which can be used to pick one of the Quine atoms; in MMK, while choice functions do exist, there isn't a choice of one particular one that an isomorphism has to preserve).

If we add a constant $q$ and the axiom $q \in q$ (so that we can distinguish the Quine atoms), and also the existence of power sets and an infinite set, the resulting theory is synonymous with MK, by a similar argument to my previous answer, except we use $q$ to remember which Quine atom was the empty set.

If, for every atom $a$, $a \subseteq x$ implies $a \subseteq y$, then $x \subseteq y$. Assume $x \nsubseteq y$, then by Supplementation, there is a $z \subseteq x$ such that $\neg z \ O \ y$. By Atomicity, there is an atom $a \subseteq z \subseteq x$, and $a \nsubseteq y$ because $\neg zOy$, contradicting the assumption that if $a \subseteq x$ then $a \subseteq y$.

This implies that the usual definition $a \subseteq b \iff \forall x (x \in a \Rightarrow x \in b)$ and your definition of $\in$ from $\subseteq$ are inverses of each other, so it does not matter whether the language is $\in$ or ($\subseteq$, $\{\}$); your mereological theory is synonymous with $\sf MMK$.

MMK and MK are not synonymous. A model of MK has no nontrivial definable automorphisms, but any model of MMK has a definable automorphism by swapping the two Quine atoms. If they were synonymous, then a model of MK would also be a model of MMK (with a different $\in$), and therefore have a nontrivial definable automorphism, which is impossible. (A similar argument might work to prove they are not bi-interpretable, but I'm not sure of the details). So the issue is that we can't distinguish the two Quine atoms. (The reason this argument didn't work in the case of ZFGC was the global choice function $C$, which can be used to pick one of the Quine atoms; in MMK, while choice functions do exist, there isn't a choice of one particular one that an isomorphism has to preserve).

If we add a constant $q$ and the axiom $q \in q$ (so that we can distinguish the Quine atoms), the resulting theory is synonymous with MK, by a similar argument to my previous answer, except we use $q$ to remember which Quine atom was the empty set.