Some context
In this answer and comments, an argument is given for why an ordered ring with unity can't have nontrivial idempotents. I'm trying to extend the argument for an ordered ring $R$ without unity (an "rng"), and I'm fairly certain it can be done as described in the comments: if $\epsilon, \delta \in R$ are nonzero idempotents such that $\epsilon\delta = \delta\epsilon = 0$, then $\epsilon(\delta-\epsilon) = -\epsilon$ and $\delta(\delta-\epsilon) = \delta$, hence $0 \leq \delta - \epsilon \leq 0$ and so $\delta = \epsilon$.
The hurdle is that this only proves that $R$ has at most one nonzero idempotent, not no nonzero idempotents. This was a problem in the ring with unity case because $1-e$ is idempotent whenever $e$ is. So, my question essentially is, is there still a problem in the "rng" case?
Question
To be more precise, if $R$ is an "rng", then we say a pair of nonzero idempotents $\epsilon, \delta \in R$ is "orthogonal" if $\epsilon\delta = \delta\epsilon = 0$. I want to know, if $R$ has a nonzero idempotent $e \in R$, under what conditions does there exist another idempotent $\delta \in R$ orthogonal to $e$?
The above work shows that no ordered "rngs" have orthogonal pairs. These posts discuss similar issues, but don't quite address orthogonality. Can we get anything more concrete?
