integral depending on a parameter

Michael has essentially answered this in his comment, but let me make this more explicit.

In fact, a stronger statement is true: If $F(z)=\int_0^1 f(t)e^{tz}\, dt$ satisfies $|F(s)|\lesssim e^{(a+\epsilon)s}$ for $s>1$ and all $\epsilon>0$ (but with possibly $\epsilon$ dependent implied constants), then $f=0$ on $[a,1]$. (This is of course extremely plausible right away, or how could there be cancellations between the various exponentials for large $s>1$?)

By splitting $0\le t\le 1$ into the two parts $[0,a+\epsilon]$ and $[a+\epsilon, 1]$, we see that the claim is equivalent to the following variant of it: If $G(z)=\int_0^b g(t)e^{tz}\, dt$ is bounded for $z=s\ge 0$, then $g\equiv 0$.

Since $G$ is of exponential type, the Phragmen-Lindelof principle applies to all sectors of opening $<\pi$, and in particular, it applies to quarter planes. Since $G$ is bounded on the imaginary axis and on $z=s\ge 0$, it is bounded on the right half plane. It is also, trivially, bounded on the left half plane. Thus $G$ is constant, and the constant is zero since $G$ is also square integrable on the imaginary axis.