As Oscar has explained in comments, with the most common definitions it's just not true that $M \sqcup N$ is exactly the same as $N \sqcup M$. But even if you were working with some version of the category of sets where they were equal, this wouldn't be a problem because tensor functor don't have to be strict. In other words, you don't know that $\mathscr{F}(M \sqcup N) = \mathscr{F}(M)\otimes \mathscr{F}(N)$, but instead only that you have isomorphisms $\mathscr{F}(M \sqcup N) \cong \mathscr{F}(M)\otimes \mathscr{F}(N)$ satisfying the natural coherence conditions. In particular, $\mathscr{F}(M \sqcup N) = \mathscr{F}(N \sqcup M)$ does not imply that $\mathscr{F}(M)\otimes \mathscr{F}(N) = \mathscr{F}(N)\otimes \mathscr{F}(M)$, but only that the latter are isomorphic in a coherent way.

As Oscar has explained in comments, with the most common definitions it's just not true that $M \sqcup N$ is exactly the same as $N \sqcup M$. But even if you were working with some version of the category of sets where they were equal, this wouldn't be a problem because tensor functors don't have to be strict. In other words, you don't know that $\mathscr{F}(M \sqcup N) = \mathscr{F}(M)\otimes \mathscr{F}(N)$, but instead only that you have isomorphisms $\mathscr{F}(M \sqcup N) \cong \mathscr{F}(M)\otimes \mathscr{F}(N)$ satisfying the natural coherence conditions. In particular, $\mathscr{F}(M \sqcup N) = \mathscr{F}(N \sqcup M)$ does not imply that $\mathscr{F}(M)\otimes \mathscr{F}(N) = \mathscr{F}(N)\otimes \mathscr{F}(M)$, but only that the latter are isomorphic in a coherent way.