Statement: Let $L_f(t) = \int_0^\infty f(x)e^{-tx} dx$. Then if $f(x) = o(e^{ax})$ for some $a > 0$, $L_f = L_g$ implies $f = g$.
Proof: By linearity of the integral, this is equivalent to showing $L_f = 0$ implies $f = 0$. Perform the substitution $x = -log(u)$ to obtain
$$
\begin{eqnarray}
L_f(t) &=& \int_0^\infty f(x)e^{-tx} dx \\
&=& \int_0^1 u^{t-1}f(-log(u)) du
\end{eqnarray}
$$
where for $t \geq a+1$, $\displaystyle \lim_{u\to 0} u^{t-1}f((-log(u)) = \lim_{x\to\infty} e^{-(t-1)x}f(x) = 0$, so this function is continuous at 0 and the above integral makes sense. Thus, if $L_f = 0$, then $\int_0^1 u^{n}f(-log(u)) du = 0$ for all sufficiently large integers $n \geq N$. Hence by linearity $\int_0^1 p(u)f(-log(u)) du = 0$ if $p$ is a polynomial which contains only terms of size $N$ or greater.
Since $f(-log(u))$ is continuous on [0, 1], by Stone-Weierstrass we can choose a sequence of polynomials $p_k$ (having terms only of size $N$ or greater) which uniformly converges to $f(-log(u))$. That is, for any $\epsilon \geq 0$, $|f(-log(u)) – p_k(u)| \leq \epsilon$ for all $u$ when $k$ is sufficiently large. Thus, on the one hand, $k$ sufficiently large implies
$$
\begin{eqnarray}
\left| \int_0^1 (f(-log(u)) – p_k(u))f(-log(u)) du \right|
&\leq& \int_0^1 \epsilon |f(-log(u))| du \\
&=& \epsilon \int_0^1 |f(-log(u))| du
\end{eqnarray}
$$
which can be made arbitrarily small. But on the other hand,
$$
\begin{eqnarray}
\int_0^1 (f(-log(u) – p_k(u))f(-log(u)) du
&=& \int_0^1 f(-log(u))^2 du + \int_0^1 p_k(u)f(-log(u)) du \\
&=& \int_0^1 f(-log(u))^2 du + 0.
\end{eqnarray}
$$
But $f(-log(u))^2$ is non-negative, and the whole integral goes to zero with $n$ (since it’s smaller than any positive $\epsilon$). The only way this is possible is if $f(-log(u)) = 0$ for all $u$ in $[0, 1]$, i.e., $f(x) = 0$ for all $x$ in $[0, \infty)$. $\blacksquare$
Notes: This can also be used to prove the uniqueness of the Fourier transform.
We needed to be able to approximate an arbitrary continuous function $f$ on $[0, 1]$ such that $f(0) = 0$ uniformly with polynomials with terms only of degree greater than or equal to some K. To do this, it suffices to be able to approximate the curve $y = x$ with those polynomials, since then you can use the approximation to $y = x$ to build up other polynomials and exploit Weierstrass approximation. To do this, set $f_n(x) = x(1 – (1 – x)^n)^K$.