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Given a (unital associative) ring $R$ and $a,b \in R, a\neq 0, b\neq 0, ab=0$, can we show that there exists an element $e \in R$ such that $e^2=e, e\ne 0, e\ne 1$?

If this is not so, does this change if we further require $R$ to be a real algebra?

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The ring $\mathbb R[X]/(X^2)$ is a real algebra with zero divisor and just trivial idempotents.

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    This answers the question as stated. Is there a nondegenracy condition that I can add to $R$ that ensures the existence of such an $e$? – 2017-02-05
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    I doubt it. Zero divisors and idemptents are somewhat orthogonal phenomena. – 2017-02-05
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    Not entirely orthogonal, since the converse holds: every nontrivial idempotent in a unital ring is a nontrivial two-sided zero divisor, so they cannot be regarded as orthogonal; this was what prompted the question. But I'll believe you on the "I doubt it". – 2017-02-05
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    That's true, but not what I meant. Idempotents are essentially relaed to splittings of modules in direct sums, and zero divisors and nilpotents are related to internal structure of modules. – 2017-02-05
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    Okay, to my untutored mind that says that the two generalize in unrelated ways, but when the module is a ring, the latter implies the former. Thanks. – 2017-02-05
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    One could say that the divisors of zero that come from the fact that a module is a direct sum are very uninteresting. They are there, sure, but among divisors of zero they are pretty irrelevant. – 2017-02-05
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A commutative ring with only trivial idempotents is called connected, and there are lots of connected rings that aren't domains.

Most of the ones appearing in the search query can be taken to be $\mathbb R$ algebras. Many of them are simply local rings that aren't domains (including the one that matches Mariano's example.)

If you're still willing to consider things that aren't $\mathbb R$ algebras, then notice the one $\mathbb Z[x]/(x^2-1)$, which has no nontrivial idempotents, isn't local, and isn't a domain.

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    I am interested in the noncommutative real algebras, but generalizations are also interesting, such as possible generalization of connectedness to the noncommutative case. – 2017-02-05
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    @qman it is easy to modify many of the examples given to be noncommutative real algebras: you just use quaternions as a base field. – 2017-02-05
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    @qman I think the generalization to noncommutative rings for "connected" is "no nontrivial central idempotents", but one may still talk about "no nontrivial idempotents." – 2017-02-05
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    So "connected" would seem to equate to what Wikipedia calls [directly irreducible](https://en.wikipedia.org/wiki/Irreducible_ring). Though I need to wrap my head around the various shades of irreducibility. – 2017-02-06
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    @qman yes, central idempotents correspond to ring decompositions into direct products. – 2017-02-06
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    It just seems that there are three terms (connected, directly irreducible and indecomposible) that seem to mean the same thing, yet this is not made clear in WP. Here I'm ignoring the distinction between a direct sum and a direct product. – 2017-02-06
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    @qman sure well, let me know if a question comes to mind. – 2017-02-06