The ordinals as a free monoid-like entity Let $\kappa$ denote a fixed but arbitrary inaccessible cardinal. Call a set $\kappa$-small iff its cardinality is strictly less than $\kappa$. By a $\kappa$-suplattice, I mean a partially ordered set whose every $\kappa$-small subset has a join. Now recall that $\mathbb{N}$ can be characterized as the (additively-denoted) monoid freely generated by the set $\{1\}$. It seems likely that the ordinal numbers can be characterized similarly. To this end, define that a $\kappa$-monarchy is a $\kappa$-suplattice $S$ equipped with an (additively-denoted) monoid structure subject to two axioms:
Axiom 0. Addition is inflationary.
$$x+y \geq x, \qquad x+y \geq y$$
Note that these can be written as equations, e.g. $(x+y) \vee y = x+y$.
Hence $x+0 \geq 0$. So $0$ is the least element of $S$. Therefore $0$ is the join identity. In other words, the identity of the monoid structure is also the identity of the $\kappa$-suplattice structure.
Axiom 1. Addition on the left commutes with joins. Explicitly:
For all $\kappa$-small sets $I$ and all $I$-indexed sequences $x$ in $S,$ and all elements $y \in S$, we have:
$$y+\left(\bigvee_{i:I} x_i \right) = \left(\bigvee_{i:I} y+x_i \right)$$

Question. It seems likely that the ordinals below $\kappa$ with their usual Cantorian operations form the free $\kappa$-monarchy on the set $\{1\}$. Is this correct?

Discussion.
Let $F_\kappa(\{1\})$ denote the $\kappa$-monarchy freely generated by $\{1\}$. Then $F_\kappa(\{1\})$ has an element that can play the role of $\omega$:
$$\omega := 1 \vee (1+1) \vee (1+1+1) \vee \cdots$$
It can be seen that $1+\omega=\omega$, using Axiom 1 followed by Axiom 0. I'm cautiously optimistic that the rest of the arithmetic will behave correctly, too. Ideas, anyone?

Let $P$ be a $\kappa$-monarchy and $p \in P$. We want to show that there is a unique morphism $f : \kappa \to P$ of $\kappa$-monarchies such that $f(1)=p$. We define $f$ via transfinite recursion as follows. We let $f(0)=0$, $f(x+1)=f(x)+p$, and for limit ordinals $x$ we let $f(x) = \sup_{y<x} f(y)$. It is clear that we have to define $f$ this way, which proves uniqueness, and we only have to check that this $f$ is a morphism of $\kappa$-monarchies. It satisfies $f(0)=0$ and is continuous by construction. In order to prove $f(x+y)=f(x)+f(y)$, we use induction on $y$. The case $y=0$ is clear. If $y=z+1$ is a successor ordinal, then $$f(x+y)=f((x+z)+1)=f(x+z)+p = (f(x)+f(z))+p$$
$$= f(x)+(f(z)+p)=f(x)+f(z+1)=f(x)+f(y).$$
If $y$ is a limit ordinal, then (since $f$ is continuous)
$f(x+y) = \sup_{z<y} f(x+z)$ and hence $$f(x+y)=\sup_{z<y} (f(x)+f(z))=f(x)+\sup_{z<y} f(z)=f(x)+f(y).$$
