Here are a few more facts which seem relevant here. Let me adopt the Boyarchenko-Drinfeld terminology from the preprint I linked to in the comments above, and use the term "closed idempotent" for what I called "well-idempotent" in the question, and "open idempoent" for "co-well-idempotent".

• The first point is that if $X$ is a closed idempotent via $r: I \to X$ and simultaneously an open idempotent via $i: X \to I$, then there are two additional coherences one should ask for: (a) $ri = 1_X$ and (b) $1_X \otimes ir = 1_{X}$. In this case, let's say that the pair $(r,i)$ exhibits $X$ as a "clopen idempotent". I'd call (a) the "splitting condition" and (b) the "stability condition".

• But we get these coherences for free: if $X$ is exhibited as a closed idempotent by $r$ and an open idempotent by $i$, then there is an $i'$ such that the pair $(r,i')$ exhibits $X$ as a clopen idempotent. The proof is similar to the proof that every equivalence can be made into an adjoint equivalence.

• Every clopen idempotent is self-dual, with unit given by $r \otimes r$ and counit given by $i \otimes i$. This is easy to show using (a) and (b).

• Conversely, if a closed idempotent has a dual (and hence is self-dual), then it is clopen.

So this was really a question about clopen idempotents. Also, everything seems to continue to work in the braided case. Two other interesting facts proved in the Boyarchencko-Drinfeld preprint:

• If $X$ is a closed idempotent, then the braiding $X \otimes X \to X \otimes X$ is equal to the identity.

• If $X$ is a closed idempotent, then the map $r$ exhibiting it as such is uniquely determined. (In particular, above we must have $i' = i$.)