The $2$-category of monoids

First of all, even having one object is not invariant under equivalence. So perhaps a monoid is actually a category with a unique isomorphism class of objects. Then the way this issue is handled in topology is to consider a different 2-category: that of pointed categories, that is, categories with a distinguished object, functors preserving that object, and natural transformations which are the identity on that object. This fixes your problem: the category of pointed functors between two pointed monoids is discrete.

Kevin Carlson answered the question, but I thought I'd add my own answer based on his, with some more details.

The ($1$-)category $\mathbf{Set}$ lives inside the $2$-category $\mathbf{Cat}$, as the full sub-$2$-category on the discrete categories. The inclusion $F:\mathbf{Set}\to\mathbf{Cat}$ has a right adjoint $U:\mathbf{Cat}\to\mathbf{Set}$ that sends a category to its set of isomorphism classes.

So a "category with one object" (or, to better respect the principle of equivalence, a "category with one isomorphism class of objects") is precisely a category $\mathcal{M}$ such that there's a bijection $1\to U\mathcal M$. Since there's at most one such bijection we could equally well say that it's a category equipped with a bijection $a:1\to U\mathcal M$. But as I said in the question, this gives a $2$-category with unwanted $2$-morphisms.

Instead, the correct definition is to look at categories equipped with a particular object to which every other object is isomorphic. An object is precisely a functor from the terminal category, and the terminal category is equivalent to $F1$. So we define a monoid to be a category $\mathcal M$ equipped with a functor $a:F1\to\mathcal M$ which corresponds to a bijection $1\to U\mathcal M$ under the isomorphism $\mathrm{Hom}(F1,\mathcal M)\cong\mathrm{Hom}(1,U\mathcal M)$ given by the adjunction.

Based on this definition, it makes sense to say that a morphism between monoids $(\mathcal M,a)\to(\mathcal N,b)$ is a functor $f:\mathcal M\to\mathcal N$ such that $f\circ a\simeq b$, and that a 2-morphism $f\to g$ is given by a natural transformation $\alpha:f\to g$ such that $(\alpha\circ f)_\bullet = g(\mathrm{id}_\bullet)$ (where $\bullet$ is the object of $1$). Of course there is in fact only one such natural transformation, so this version of $\mathbf{Mon}$ is indeed a $1$-category.

The sort of definition we gave above is actually quite common in mathematics. Two similar definitions arise from the usual adjunction between $\mathbf{Set}$ and $\mathbf{Vect}$. A basis $S$ of a vector space $V$ is precisely a function $f:S\to UV$ that the corresponding function $FS\to V$ is an isomorphism. Dually a vector space structure $V$ on a set $S$ is a function $FS\to V$ such that the corresponding function $S\to UV$ is a bijection.

By analogy, we could say that a monoid is not a "category with one object" but rather a "category structure on the set with one element". This gives some intuition for why monoids only form a $1$-category. Categories naturally form $2$-categories, but structures based on sets are only sophisticated enough to form $1$-categories.

In fact, I believe that if we allow any set $S$ in place of $1$ in the above definition of a monoid then we get a definition of the $1$-category of categories. So we could also define monoids by first passing to this $1$-category, and then looking at the "categories with one object" within it.