Several conditions are equivalent to absolutely flat.
Assume (1), whence any quotient ring R/H in R is a flat R module. Tensor this with 0 → H → R → R/H → 0. Use the quotient ring formula. Naturally R becomes R/H, but R/H also becomes R/H. Descend into the algebra of the tensor product, and imagine x,y in R cross R/H becoming x,y in R/H cross R/H. For this to be 0, some factor of x or y must lie in H, but then we could pass this factor of x over to y, and x,y would be 0 in R×(R/H). In other words, R×(R/H) → (R/H)×(R/H) has no kernel, and is an isomorphism. Since the sequence is exact, H×(R/H), which is H/H2, is equal to 0. Therefore H2 = H, and H is an idempotent ideal.
Now assume H is generated by x. Thus xxy = x for some y in R. Set e = xy to find an idempotent that generates H.
If H has two generators, replace them with their idempotent equivalents e and f. Now e+f-ef generates H and is idempotent. By induction, every finitely generated ideal is generated by an idempotent. Equivalently, any proper finitely generated ideal is a summand of R. Let e generate the ideal, and let f = 1-e generate the "other" summand.
Finally go from (3) back to (1). Let H be a finitely generated ideal of R, with G as the "other" summand. Thus 0 → H → R → G → 0 is split exact. Tensor with any module M and the result is split exact. Thus tor(R/H,M) = 0 for every finitely generated ideal H. This implies M is flat, hence R is absolutely flat.
As a special case, every boolean ring is absolutely flat.
Interestingly, these rings consist of units and zero divisors. The ideal generated by x equals the ideal generated by x2, so for some y, xxy = x, whence x*(1-xy) = 0. If x is not 0 and is not a 0 divisor then x is invertible.
conversely, every field is absolutely flat, since the only nontrivial ideal is generated by 1, which is idempotent. Another proof: every module over a field is free, hence flat.