It is quite easy to show that if $\mu$ is positive finite Borel measure on, say $[0,1]$, and for all $n \in \mathbb{N}$

$$\int_{[0,1]} e^{-nx}\mu(dx)=0$$

holds true, then $\mu=0$. Does this still hold if $\mu$ is signed finite Borel measure?

For positive measure, by applying Holder's inequality it can be shown that $\int_{[0,1]} e^{-rx} \mu(dx)=0$ holds for all positive rational $r$, hence by DCT holds for all $r\geq 0$. By taking differentiation under the integral sign and set $t=0$ we find $$\int_{[0,1]}f(x)\mu(dx)=0$$ holds for all polynomial $f(\cdot)$ and hence for continuous $f(\cdot)$ by Weistrass's approximation theorem. Now Lusin's theorem for Radon measure implies the above equality also holds for measurable $f(\cdot)$, especially for indicating functions so we reach the conclusion.

But for signed measure, it seems Holder's inequality does not work out so I tried to show $\int_{[0,1]}e^{-nx}\mu_+(dx)=0$ and $\int_{[0,1]}e^{-nx}\mu_{-}(dx)=0$ where $\mu=\mu_+-\mu_-$ is the Hahn decomposition. But it did not work out so well.

[Possible solution]Below is one possible solution I gave.

Let $T:\mathbb{R}\supset A \to B\subset \mathbb{R}$ be diffeomorphism,($\mathcal{C}^1$-diffeomorphism seems to be sufficient but smoothness isn't really what matters here), then $$\int_A f(x) \mu(dx)=\int_B f\circ T^{-1} (y) (\mu\circ T^{-1})(dy)$$ holds for all $f$ that is $\mu$-measurable and $\mu\circ T^{-1}$ is the image measure induced naturally by $T$. Now let $T={x \mapsto e^{-nx}}$, and $A=[0,1]$, then $B=[e^{-1},1]$, and $$\int_{[e^{-1},1]}x^n \tilde{\mu}(dx)=0$$ where $\tilde{\mu}=\mu\circ T^{-1}$. Now the conclusion follows immediately by similar argument as stated above.