 Solution

A decidable $B$ has injective action maps. First prove that evaluation at $1$ distinguishes equivariant functions. Suppose $f(1,a)=g(1,a)$ for every $a$. Fix $m\in M$ and choose $p,q$ with $pmq=p$. For each $a$, equivariance givesThe same equation holds for $g$, so these values agree. Cancel the injective action of $p$ on $B$ to obtain $f(m,a)=g(m,a)$. Hence $f=g$. This proves the hinted contrapositive and, more precisely, injectivity of the trace map $f\mapsto (a\mapsto f(1,a))$.

Now suppose $m\cdot f=m\cdot g$ in the exponential. Evaluating this equality at $(1,m\cdot a)$ givesCancel the action of $m$ on $B$. The trace maps agree, so the preceding argument gives $f=g$. Every action map on $B^A$ is therefore injective. By the introductory criterion, $B^A$ is decidable whenever $B$ is under the specified monoid condition. Neither cancellation in $M$ nor injectivity of its action on $A$ is assumed.