There exists an element of the magma c such that for all x: $ x*x=c $
The consequence of this is that the elements on the diagonal of the Cayley table are all the same, e.a:
$ * = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 1 \end{bmatrix} $
For all x in this magma: $ x*x=1 $
$c$ does not have to be $x$ (making it idempotent), or the identity element of the magma.
Just a few months of mumbling, since I first read this question, and I'm ready to give you an answer. First things first.
A magma $(G,\ast)$ is said to be idempotent if $x\ast x=x$ for all $x\in G$. This is the case of elements of $G$ pairwise distinct on the diagonal of the Cayley table: $$\begin{array}{c|ccc} \ast & a & b & c & \ldots\\ \hline a & a & & & \\ b & & b & & \\ c & & & c & \\ \vdots & & & & \ddots\\ \end{array}$$
Instead, in literature we find two definitions for unipotence of a magma: $$x\ast x=e\qquad \forall x\in G,\tag{1}$$ where $e$ is the identity element of $G$ (so, we are assuming already that $G$ has identity element), and: $$x\ast x=y\ast y\qquad \forall x,y\in G.\tag{2}$$ The condition in line 1 of your question ($x\ast x=c$ for all $x$, for some generic $c$) is equivalent to definition (2). You already provided an example of magma without identity where the less restrictive definition (2) holds. We can assume this latter as the definition of unipotence for an arbitrary magma while in a magma with identity $e$, hence in all monoids, definition (2) implies definition (1) since it must be $e\ast e=e$.
If (1) holds, but also if only (2) holds, then the elements on the diagonal of the Cayley table are all the same, as you mentioned.