This is just a rough outline and may be wrong, but I thought I'd toss it out and see if I could learn something from my likely mistakes.
Let $I$ be a model of PA and $M$ a model of MA. Let $A=I\times M$ be the model of PA defined by taking the Cartesian product of the two sets and defining $S(i,m)=(i,S(m))$ if $S(m)\ne 0$, else $(S(i),0)$.
To check that $A$ is a model of PA as claimed, the main issue is the verification of induction. Suppose that the hypotheses of induction hold for some set K. If K includes $(i,0)$, then $(i,m)$ is included for all $m$ because induction is a hypothesis of MA. Also, if $(i,0)$ is in K, then so is $(i+1,0)$, since one of the axioms of MA is that there is an $m$ whose successor is 0, and we've already found that all $(i,m)$ are in K. This establishes that induction holds for $A$.
I'm pretty sure that
As with any model of PA, the definition of S implies uniquely defined operations of addition and multiplication. Under this definition of addition, any member $(i,m)+(i',m')$ (i,m)$ of A can be written as $(i,0)+(0,m)$, and this form is unique. For any two elements of A, we then have $(i,m)+(i',m')=(i+i',0)+(0,m)+(0,m')$, which is of the form $(\ldots,m+m')$.
Suppose that Russell Easterly's conjecture fails, so there is no nonzero $m$ in $M$ such that $m+m=0$ or $m+m=1$. Then there is no nonzero $a$ in $A$ such that $a+a$ equals $(1,0)$ or $(1,1)$. But it's a theorem in PA that for any number $x$, either $x$ or $S(x)$ can be expressed as $y+y$, so we have a contradiction.
This establishes the "or" part, and Ashutosh's comment shows that it's not just or but xor.
I think this also leads to the fact that a model of MA is always isomorphic to a model of PA modulo one of its elements. Since nonstandard models of arithmetic are pretty thoroughly studied, I think that this also characterizes the possible nonstandard models of MA.