a partial answer:

if such bijections x→xu and x→ux exists for some element u, then for some element x:

xu = u, and since for ANY y in S we have y = ua (for some a): xy = x(ua) = (xu)a = ua = y. thus x is a left-identity for S.

similarly, there must be some z in S with uz = u, and thus (writing y = bu), yz = (bu)z = b(uz) = bu = y, so z is a right-identity for S.

but then x = xz = z, so S possesses an identity element, which we can call e.

so in THAT case, we have xu = e and uy = e for some x,y in S, so u is indeed a unit (and x = y: x = xe = x(uy) = (xu)y = ey = y).

note that above, we only require that x→ux, x→xu be surjective to show that a right- or left-identity exists, uniqueness of this identity follows (for one of left- or right-) if one of those maps is injective as well (if S is finite, then surjective = bijective) (if x→ux is injective, then we have a unique right-identity, if x→xu is injective, we have a unique left-identity).

i am unsure of how you would define a unit without the presence of a two-sided identity.