There is a version of Mereology in which there is no Bottom atom, and yet it is synonymous with the above theory, and if I'm correct about the above theory $\sf M$ being synonymous with MK-Foundation-Choice, then that answers this question to the positive.

The idea is to modify the underlying mereology as to have a kind of alien atom mereologically speaking, that is an object that doesn't overlap with any other object! This was done in a [prior][1] posting. Now, to enable that here, we need to adjust composition as to forbid composing objects using that atom. Now, we use this atom to be the non-labeling atom. Here is the exposition of it:

**Language:** first order logic with equality, with primitives of $\subseteq$ standing for Part-hood, and $\{ * \}$ standing for the singleton function which is a partial unary function. The axioms are:

***Mereological Axioms:***

***Define:*** $atom(X) \iff \forall Y \subseteq X \, (Y = X)$

***M1*** $\textbf{Atomism: } \forall \ atom \ X \ (  X \subseteq Y \to   X \subseteq Z) \leftrightarrow Y  \subseteq Z$


***Define:*** $ X  \subsetneq Y \iff X  \subseteq Y \land X \neq Y $

 
***Define:*** $atom^*(X) \iff atom(X) \land \exists Y: X \subsetneq Y $
 
***M2*** $\textbf{Extensionality: }\forall \ atom^* \ X \, (X \subseteq A \leftrightarrow X \subseteq B ) \to A=B$

***Define:*** $X \ \mathcal O \ Y \iff \exists Z: Z \subseteq X \land Z \subseteq Y$

***M3*** $\textbf{Alien: }\exists X \forall Y: Y \neq X \to \neg X \ \mathcal O \ Y$

***Define:*** $X = \varnothing \iff \forall Y: Y \neq X \to \neg X \ \mathcal O \ Y$

***M4*** $\textbf{Composition: }   \exists X \forall \ atom^* \ Y \ (  Y \subseteq X \leftrightarrow  \psi( Y )); \text{ if } X \text { is not free in } \psi $

***Define:*** $X = \bigl[Y \mid \psi \bigr] \iff \forall \ atom^* \ Y \ (  Y \subseteq X \leftrightarrow  \psi( Y )) $

***Labeling Axioms***

***L1*** $\textbf{Labeling: } \{X\}=\{Y\} \to X=Y$

***L2*** $\textbf{Purity: } \exists X \ (\{X\} =Y)  \leftrightarrow atom^*(Y)   $

***L3*** $\textbf{Replacement: } \text  { if } \psi(X,Y) \text { is a formula, then: } \\ \forall X \neg \exists^{>1} Y: \psi(X,Y), \land   \\ B= \bigl[   K \mid K \subseteq Y \land \exists V (V=\{Y\} \lor Y=\{V\}) \land \exists X  \subseteq A \ :\psi(X,Y)\bigr] \land \\\exists Z: \{A\}=Z \\\to\\\exists Z: \{B\}=Z $

***L4*** $\textbf{Infinity: } \exists I  \exists X:   \forall \ atom  \ Y \subseteq X \bigl(\exists Z  \subseteq X: Z=\{Y\} \bigr) \land \{X\}=I  $


Now to interpret $\sf M$ in this theory, we take the domain to be all objects here, so there is no restriction, we keep the same equality relation, we re-define $P$ as 

$X \ P \ Y \equiv_{def}  X \subseteq Y \lor X=\varnothing $

, and of course define $\mathfrak L$ as:

$\mathfrak L(X)=Y \equiv_{def} Y = \{X\}$

For the opposite direction, we work in $\sf M$, keep the same domain, and equality relations, re-define $\subseteq$ as:

$ X \subseteq Y \equiv_{def} X \ P \ Y \land X \neq \varnothing$

, and of course define the singleton function as:

$\{X\}=Y \equiv_{def} \mathfrak L(X)=Y$

We'll use the equality relation as our needed  i-isomorphisms in both directions. $\small \square$

As regards the system $\sf M-Bottom$ mentioned in the question, I think it is not bi-interpretable with MK-Foundation-Choice, since it would be haunted by the same argument present in a [prior posting][2]. 


  [1]: https://mathoverflow.net/q/412970/95347
  [2]: https://mathoverflow.net/a/459098/95347