As physical scientists we should look at the numbers;
#H_2SO_(4(aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + HSO_(4(aq))^(-)#
The very important thing to realize here is that this equilibrium lies so far to the right that this reaction can be considered as going to completion.
#H_2SO_(4(aq)) + H_2O_((l)) -> H_3O_((aq))^(+) + HSO_(4(aq))^(-)#
This means that the resulting solution will contain no
So, if you add sulfuric acid to water, you can expect all the sulfuric acid molecules to donate one proton and dissociate into hydrogen sulfate and hydronium.
This implies that the hydrogen sulfate anion simply cannot accept a proton, i.e. act as a base, to reform sulfuric acid because the aforementioned equilibrium will not allow for the reverse reaction to take place.
#HSO_(4(aq))^(-) + H_2O_((l)) color(red)(cancel(color(black)(->))) H_2SO_(4(aq)) + OH_((aq))^(-)#