It's rather easy to notice that the operation of join of categories reproduces the ordinal sum once restricted to act on (iso classes of) well-ordered set; it's rather easy to see that $\alpha\star [0]$ (as a category) equals $\alpha+1$ (as number).

And in fact there is more ([Jec, p. 16]): the transfinite sequence $s\colon \mathbf{Ord}_{<\alpha}\to X\colon \langle a_\xi\mid \xi < \alpha\rangle$, extended with value $x_0$ on the successor $\alpha+1=\alpha^+$, corresponds exactly to the operation which, given a category $\cal C$, adds a (strained) terminal object $*$ defining $\mathcal C^\rhd = \mathcal C\star [0]$($=\mathcal C\coprod\{*\}$ with a unique arrow between any $C\in\cal C$ and $*$), and given a functor $F\colon \mathcal C\to \cal X$, it is extended with value $X_0\in\cal X$ on the unique object $*$ of $[0]$, coinciding with $F$ on $\cal C$.

Now: $\bf Cat$ endowed with the bifunctor $\star$ becomes monoidal [Joy, pp. 26-27] and the structure is closed on both sides (albeit not biclosed; the $\star$ is not symmetric)

Should this give a characterization of ordinals as *some sort of cocompletion of the free $\star$-monoidal category* on one generator? Can something non trivial be desumed by this fact?

**Edit:** in fact as it is stated the question is unclear. You must first complete, and then close under $\star[0]$ (I was confused, and maybe I still am, by the fact that $\star[0]$ commutes with colimits).

[Jec] Jech, Thomas. *Set theory*. Vol. 79. New York: Academic press, 1978.