You cannot instruct the H atom to be in a particular charge state, since the Mulliken population of each atom will be determined by the self-consistent state of the whole system. Thus the relevant quantity to compare to is not the isolated H atom, but the self-consistent state of the H atom in the given atomic environment. I am sure (or at least I very much hope) that if you compare the 3 calculations with +15 eV, 0 eV, and -15 eV, you will see a net additional positive/negative charge on the H atom in the two cases with a finite shift, compared to no shift.