Yes. Call your monad $$T$$. Then $$T(S)=\sqcup_{n\in \mathbb{N}}\mathrm{Set}(n,S)$$. That is, a multiset in $$S$$ is just a map to $$S$$ from a finite set. Thus the finitarity of $$T$$ follows from the finite presentability of each $$n$$, and the fact that finitary functors $$\mathrm{Set}\to\mathrm{Set}$$ are closed under colimits. Indeed, if each $$F_i$$ is finitary, $$I$$ is any category such that the colimit exists (e.g. $$I$$ is small), and $$J$$ is filtered, then $$(\mathrm{colim}_I F_i)(\mathrm{colim}_J S_j) =\mathrm{colim}_I\mathrm{colim}_J F_i S_j=\mathrm{colim}_J((\mathrm{colim}_I F_i)(S_j)$$ using the facts that colimits in functor categories are pointwise, colimits commute with colimits, and the assumptions on $$F_i$$.
It's not correct that $$T$$'s algebras are free commutative monoids. A $$T$$-algebra is a summation map $$TS\to S$$ that turns any finite multiset in $$S$$ into an element of $$S$$ in a way coherent with the structure of $$T$$. So it's an arbitrary commutative monoid. This gives a more general way to see that $$T$$ is finitary: it's the monad for an equational algebraic theory.