But it's possible that the compiler is smart enough to optimize the step() version down to the same code as the conditional version. If true, that still wouldn't justify using step(), but it would mean that the step() version isn't "wasting two multiplications and one or two additions" as the post says.
(I don't know enough about GPU compilers to say whether they implement such an optimization, but if step() abuse is as popular as the post says, then they probably should.)
Okay but how does this help the reader? If the worse code happens to optimize to the same thing it's still awful and you get no benefits. It's likely not to optimize down unless you have fast-math enabled because the extra float ops have to be preserved to be IEEE754 compliant
Fragment and vertex shaders generally don't target strict IEEE754 compliance by default. Transforming a * (b ? 1.0 : 0.0) into b ? a : 0.0 is absolutely something you can expect a shader compiler to do - that only requires assuming a is not NaN.
(I don't know enough about GPU compilers to say whether they implement such an optimization, but if step() abuse is as popular as the post says, then they probably should.)