Your "trivial" examples all resulted from direct sum decompositions of the ring R. By asking for examples without idempotents, you are asking for rings that do not have direct sum decompositions. In a noncommutative ring R, the corresponding would be a ring that has no central idempotents. I can provide a noncommutative example that is "in-between," so that it has nontrivial idempotents, but no nontrivial central idempotents.

For an ideal J in a noncommutative (read: not-necessarily-commutative) ring R, there is a way to reformulate when the right R-module R/J is flat. In T.Y. Lam's Lectures on Modules and Rings, Proposition 4.14 implies that R/J is right flat if and only if, for every left ideal _RL ⊆ R,

J ∩ L = JL.

(Notice that this provides an alternative way to verify that for such J, J² = J.)

Now given a field k (or even a division ring!), let V be a (right) vector space of countably infinite dimension, and let R = End_k(V), acting on V from the left. This ring has many idempotents, corresponding to direct sum decompositions of V. One can show that R has precisely three ideals, namely 0, R, and the ideal J consisting of endomorphisms of finite rank (see Exercises 3.15-3.16 of Lam's Exercises in Classical Ring Theory). In particular, R does not decompose as the direct sum of two nontrivial subrings. Let f be any finite-rank element of R, and let p in R be a projection of V onto the image of f. Certainly f = pf ∈ Jf. This makes it easy to show that J satisfies J ∩ L = JL for every left ideal L of R, and it follows that R/J is flat.

Your "trivial" examples all resulted from direct sum decompositions of the ring $R$. By asking for examples without idempotents, you are asking for rings that do not have direct sum decompositions. In a noncommutative ring $R$, the corresponding would be a ring that has no central idempotents. I can provide a noncommutative example that is "in-between," so that it has nontrivial idempotents, but no nontrivial central idempotents.

For an ideal $J$ in a noncommutative (read: not-necessarily-commutative) ring $R$, there is a way to reformulate when the right $R$-module $R/J$ is flat. In T.Y. Lam's Lectures on Modules and Rings, Proposition 4.14 implies that $R/J$ is right flat if and only if, for every left ideal $_RL \subseteq R$0, $$ J \cap L = JL. $$ (Notice that this provides an alternative way to verify that for such $J$, $J^2 = J$.)

Now given a field $k$ (or even a division ring!), let $V$ be a (right) vector space of countably infinite dimension, and let $R = \operatorname{End}_k(V)$, acting on $V$ from the left. This ring has many idempotents, corresponding to direct sum decompositions of $V$. One can show that $R$ has precisely three ideals, namely $0$, $R$, and the ideal $J$ consisting of endomorphisms of finite rank (see Exercises 3.15-3.16 of Lam's Exercises in Classical Ring Theory). In particular, $R$ does not decompose as the direct sum of two nontrivial subrings. Let $f$ be any finite-rank element of $R$, and let $p$ in $R$ be a projection of $V$ onto the image of $f$. Certainly $f = pf \in Jf$. This makes it easy to show that $J$ satisfies $J \cap L = JL$ for every left ideal $L$ of $R$, and it follows that $R/J$ is flat.