Let $\Omega=[0,\infty)$, and $\mu$ and $\nu$ be measures including $\{[a,b)|0\leq a \leq b \leq \infty\}$.
Suppose for all natural numbers $n$, their Laplace transformations are coincide, they are same measure. $F(z):=\int e^{-\frac{x}{z}}d\mu (x), G(z):=\int e^{-\frac{x}{z}}d\nu (x)$, then $F(\frac{1}{n})=G(\frac{1}{n})$ and $\lim \frac{1}{n} =0$. So if $F$ and $G$ are holomorophic on domain including $0$ ,then $F=G$. But I can't prove holomorophism.
On
Your $F$ and $G$ are holomorphic in $\Omega=\{\Re z>0\}$ (a clean way to see this is to first show they're continuous and then apply Morera's Theorem). Alas $0$ is just a boundary point of $\Omega$, so the argument doesn't quite work, or at least it's not quite that simple.
One way to fix it is to show that $F$ and $G$ are bounded and to note that the sequence $1/n$ violates the Blaschke condition regarding zero sets of bounded holomorphic functions in a half-plane.
That uses moderately "advanced" complex analysis. There's a simple real-variables proof: Define two new measures by $d\mu'(x)=e^{-x}d\mu(x)$ and $d\nu'(x)=e^{-x}d\nu(x)$. So $\mu'$ and $\nu'$ satisfy the same hypothesis, except for even $n$. Stone-Weierstrass shows that $\int\phi d\mu'=\int\phi d\nu'$ for every $\phi\in C_0([0,\infty))$; hence $\mu'=\nu'$.
$f(z)=\int e^{-zx} d\mu(x)$ and $g(z)=\int e^{-zx} d\nu(x)$ are holomorphic in $\Omega \equiv \{z \in \mathbb C :\Re z>0\}$. [you can prove this by directly showing differentiability]. The map $z \to \frac 1 z$ maps $\Omega$ onto itself. Since $f(\frac 1 z)$ and $g(\frac 1 z)$ coincide at the points $\frac 1 n, n \geq 1$, $n$ odd they are equal on $\Omega$. Hence $\int e^{-tx} d\mu(x)=\int e^{-tx} d\mu(x)$ for all $t>0$ and this implies $\mu =\nu$.