Is it possible to give a 'categorical' criterion for when an arbitrary element $m$ of a monoid $M$ is invertible, by which I mean, a criterion that ideally only refers to monoid homomorphisms and does not explicitly refer to the monoid structure?
For example, it is (obviously) true that an element $m$ of a monoid $M$ is invertible iff its image under any monoid homomorphism is invertible. But here the right-hand criterion still involves the notion of invertibility in a monoid. Is there a right-hand criterion that does not make use of the explicit monoid structure? I'm guessing the answer will be 'no', but I just wanted to be sure.
UPDATE: The application I have in mind is the following: I want to determine, for an arbitrary algebraic/equational theory $T$, whether I can create a suitable definition of an 'invertible element' of a model $M$ of $T$. So if I could determine a criterion for when an element of a monoid is invertible, which could be stated in a 'categorical' or universal algebraic way that could be applied to any algebraic theory, then this would likely give me what I want.
Without having your actual application in mind, my first reaction is as follows.
"Categorically", we tend to identify elements $m \in M$ with homomorphism $\mathbb{N} \to M$. And we can identify invertible elements of $M$ with homomorphisms $\mathbb{Z} \to M$.
So, an element $m : \mathbb{N} \to M$ is invertible if and only if it can be written as a composite of the form $\mathbb{N} \to \mathbb{Z} \to M $, where the first map is the canonical inclusion.
My next reaction, again still not knowing if its suitable for your application, is to pick out the functor $\mathrm{Core}$ (the subgroup of invertible units) in an abstract way.
If $U : \mathbf{Groups} \to \mathbf{Monoids}$ is the forgetful functor, then I believe there is an adjunction $\hom_{\mathbf{Monoids}}(U(G), M) \cong \hom_{\mathbf{Groups}}(G, \mathrm{Core}(M))$.
Thus, you can identify the functor $\mathrm{Core}$ as being the right adjoint to $U$, and the counit $U \mathrm{Core}(M) \to M$ picks out its submonoid of invertible elements.
And, of course, an element $m \in M$ is invertible if and only if it lies in this submonoid.