I think the user Av's answer (using Liouville's theorem) is correct, and a very simple argument, though it does use the fairly hefty machinery that goes into "a bounded analytic function is constant".
EDIT: as tome kindly points out, e^x f(x) need not remain bounded as x goes to -infinity, or indeed any other direction of infinity on the complex plane.
This problem is interesting but not quite air-tight. You could just take the first sequence and set some element to 0 and you've got a new sequence that satisfies the property. Is he actually asking whether there's not another A between 0 and 1 such that the series converges for a_n = C(-A)^n?
No, you couldn't. Suppose $f(x)$ decays like $O(e^{-x})$ and $a_0$ is the constant term in its power series expansion. Then $g(x)=f(x)-a_0$ decays like $O(e^{-x}$ if and only if $f-g$ decays like $O(e^{-x})$, which in turn is equivalent to the condition that $a_0=0$. In other words, you either have $a_0=0$ in the first place (but then you have a circular argument, not a genuine proof), or $g(x)=f(x)-a$ not decaying like $O(e^{-x})$.
I was thinking that any function like f(x)=e^(-x(1+o(1/x^2)) would work, provided that it can be expended into an infinite series (say, f(x)=e^(-x(1+e^(-x))) )
EDIT: as tome kindly points out, e^x f(x) need not remain bounded as x goes to -infinity, or indeed any other direction of infinity on the complex plane.