Note: I originally posted an answer claiming the opposite, and then deleted it because it was wrong. I have since reworked it but my cardinal arithmetic is still all wrong. I have made this post community wiki in case anyone is interested in clearing up what's going on in the infinite case.
The following is probably all wrong although I suspect the conjecture is true in the infinite case.
This doesn't directly answer your question, but you may be interested to know that the conjecture is true in the infinite case. In fact, for every infinite $\kappa$ there are $2^\kappa$ categories with $<\kappa$ morphisms, up to equivalence or isomorphism, $2^\kappa$ of which are monoids.
On the one hand, $2^\kappa$ is an obvious upper bound on the number of categories with $<\kappa$ morphisms: since composition is a partial binary operation on the set of morphisms, there are at most $\sum_{\lambda<\kappa} 2^{\lambda\times \lambda \times \lambda} \leq \kappa 2^\kappa = 2^\kappa$ category structures.
On the other hand, it's easy to find $\kappa$-many pairwise-inequivalent connected categories with $<\kappa$ morphisms: for instance, it suffices to consider the oridnals. By taking disjoint unions of these, we get $2^\kappa$ pairwise-inequivalent categories.
We can tweak this construction to get $2^\kappa$ pairwise-inequivalent monoids. For each $S\subseteq \kappa$, we can adjoin a bottom element to $\coprod_{\lambda \in S} \lambda$ to get a semilattice $(\coprod_{\lambda \in S} \lambda)_{\bot}$, and then consider the corresponding monoid under $\vee$. We obtain $2^\kappa$ monoids this way by taking different $S$, and they are still pairwise nonisomorphic as monoids, because a monoid isomoprhism entails a semilattice isomorphism, which entails an isomorphism of the original disjoint-union categories, which are clearly pairwise inequivalent.