Why does Mathematica not seem to know that $i^x$ cannot be equal to $0$?

We can actually 'teach' Simplify using the option TransformationFunctions.

dropPows[Power[_?NumericQ, _] a_ == 0] := a == 0

Simplify[I^x a == 0, TransformationFunctions -> {Automatic, dropPows}]
(* a == 0 *)

Maybe this is a difference in versions (9.0.1), but when I do:

FullSimplify[2^x a == 0, x \[Element] Reals]
a == 0

Replacing 2 with I gives the same answer

FullSimplify[I^x a == 0, x \[Element] Reals]
a == 0

Counterexample, of arguable importance:

FunctionExpand[I^x] /. x -> DirectedInfinity[I]
(*  0  *)

Which shows why @bill s's idea of restricting x to a number might help simplify the expression. I don't know if FullSimplify was necessary back then, but in V11 the following works:

Simplify[I^x a == 0, x ∈ Complexes]
(*  a == 0  *)