The ionization energy of H₂⁺ molecule ion
I would like to clarify the matter since from OP's own answer I have a feeling he/she does not quite understand it.

So, first of all, in general the ionisation energy has to be calculated as the difference in total energy and not just the electronic one.

Secondly, for molecules there are two types ionization energies: adiabatic and vertical:


  The adiabatic ionisation energy refers to the formation of the
  molecular ion in its ground vibrational state and the vertical
  ionization energy applies to the transition to the molecular ion
  without change in geometry. (From IUPAC Gold Book)


Note that in accordance with the Franck–Condon principle the most probable transition is the vertical one. Consequently, usually, when we say "ionisation energy" we actually mean the "vertical ionisation energy".

Finally, if the total energy is understood in the clamped nuclei approximation, i.e. as the electronic energy $E_{\mathrm{e}}$ together with the nuclear repulsion energy $V_{\mathrm{nn}}$, then for the vertical ionisation energy one could indeed use the difference in electronic energy rather than in total one simply because the nuclear repulsion energy stays the same. Mathematically, $V'_{\mathrm{nn}} = V_{\mathrm{nn}}$ since geometry is not changing, and thus,
$$
  E_{\mathrm{i}}
  =
  U' - U
  =
  (E'_{\mathrm{e}} + V'_{\mathrm{nn}})
  -
  (E_{\mathrm{e}} + V_{\mathrm{nn}})
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}} \, ,
$$
where primed quantities refers to the resulting molecular specie and non-primed are that of the original molecular specie. 

So, yes, for the vertical ionisation energy in the clamped nuclei approximation one can use the difference in electronic energy. And one can do so regardless of what are the molecular species in question. For the specific case of the vertical ionisation of $\ce{H2+}$ the electronic energy of the resulting $\ce{H2^2+}$ specie is obviously zero, so we specifically get what OP suggested
$$
  E_{\mathrm{i}}
  =
  E'_{\mathrm{e}} - E_{\mathrm{e}}
  =
  0 - E_{\mathrm{e}}
  =
  - E_{\mathrm{e}} \, .
$$
However, this is just a trivial simplification of a general formula for the vertical ionisation energy for the case of a zero-electron resulting specie.