No, empty A is critical to the counterexample. In your example, g(x) = 1 is a left inverse.

The point is you either send an element of the codomain to its (unique by injectivity) preimage if it's in the image, or to an arbitrary element of A if it's not, and that's a left inverse. But then if B has an element, A needs one for you to pick your arbitrary target.

In a sense, your claim that the problem is a smaller domain than codomain does contribute though; if f is also surjective, then this case can't happen, so bijective iff invertible (the empty function is vacuously bijective and its own inverse).

Oh, oops, you're right. Sorry.