Under what choice assumptions is there a monoid structure on every set?

Every non-empty set admits a commutative monoid structure:

1. The 1-element set is a monoid in a unique way.
2. If $X$ has at least two distinct elements, say $0$ and $1$, then we can make $X$ into a commutative monoid as follows: $$x \cdot y = \begin{cases} x & \text{if } y = 1 \\ y & \text{if } x = 1 \\ 0 & \text{otherwise} \end{cases}$$ Of course, $X$ is not a cancellative monoid in this case.

If you're willing to think about semigroups rather than monoids, then some further comments are in order.

First comment. Every set $X$ carries two canonical semigroup structures:

• The "left zero" semigroup: $xy=x$ for all $x,y \in X$.
• The "right zero" semigroup: $xy=y$ for all $x,y \in X$.

This gives two non-equivalent functors

$$\mathbf{SemiGrp} \leftarrow \mathbf{Set}.$$

Obviously, both are sections of the forgetful functor

$$\mathbf{SemiGrp} \rightarrow \mathbf{Set}.$$

Second comment. By a pointed semigroup, I mean a semigroup $X$ together with a distinguished absorbing element $\bot$. This can be equivalently defined as a semigroup object in the monoidal category of pointed sets with smash product.

Its interesting to observe that every pointed set $(X,\bot)$ can be made into a pointed semigroup in the following ways:

• The "Zhen Lin" semigroup: we declare that $xy=\bot$ for all $x,y \in X$.
• The "equality" semigroup: we declare that $x^2=x$ for all $x \in X,$ and $xy=\bot$ for all distinct $x,y \in X$.

This gives two non-equivalent functors

$$\mathbf{\bot SemiGrp} \leftarrow \mathbf{\bot Set}.$$

Obviously, both are sections of the forgetful functor

$$\mathbf{\bot SemiGrp} \rightarrow \mathbf{\bot Set}.$$

Third comment.

What we're talking about here is really a kind of "categorified idempotency."

Let me explain.

If $x$ is an element of some monoid, then $x$ is said to be idempotent iff $x=x^2$. It makes sense to write $\mathrm{isIdem}(x)$ for the statement $x=x^2$. Okay, but what happens if we're given an object $X$ living in a monoidal category? Well, we could choose to define that $\mathrm{isIdem}(X)$ means "the set of all isomorphisms $X \leftarrow X \otimes X$." But such isomorphisms seem not to arise very often in practice. Another option would be to defined that $\mathrm{isIdem}(X)$ means "the set of all morphisms $X \leftarrow X \otimes X$." So basically, its the set of all ways of making $X$ into a magma. But darn it, magmas are just so uselessly general! So to make the concept more useful, lets define that $\mathrm{isIdem}(X)$ is the set of all associative morphisms $f : X \leftarrow X \otimes X$. In other words, its the collection of all ways of making $X$ into a semigroup.

Under the philosophy of propositions as types and in light of the above paragraph, we can think of the proposition "$X$ is idempotent" as denoting the collection of all ways of making the object $X$ into a semigroup. Hence if $\mathbf{C}$ is a monoidal category, we can think of the proposition "every object of $\mathbf{C}$ is idempotent" as denoting the collection of all sections of the forgetful functor $$\mathbf{C} \leftarrow \mathrm{Semi}(\mathbf{C}),$$ which are thought of as "proofs" for this statement. Hence, the first comment really consisted of defining two non-equivalent "proofs" that "every object of $(\mathbf{Set},\times,1)$ is idempotent," and the second comment really consisted of defining two non-equivalent "proofs" that "every object of $(\bot\mathbf{Set},\otimes,\{0,\bot\})$ is idempotent."