My professor says that the Laplace transform of a nonnegative RV uniquely determines the RV up to distributional equality among all nonnegative RVs. He says one can argue this by appealing to a fact I already know, which is that the distribution of an RV is determined by its characteristic function.
I don't see how to attack the problem this way, nor any other way. If the RVs were bounded, I think I'd know what to do. The ch.f would then have an analytic continuation to the entire complex plane, and a perpendicular slice of the function gives the Fourier transform, so therefore both must have the same ch. f.
Let $X$ and $Y$ be non-negative random variables which have the same Laplace transform. Define $$f(z):=\int_{\Bbb R}e^{-tz}d\mu_X(t)-\int_{\Bbb R}e^{-tz}d\mu_Y(t),$$ where $z\in\ U:=\{z=x+iy,x>0\}$. Then $f$ is analytic on the connected open set $U$, and is $0$ on the non-discrete subset $\{x,x>0\}\subset U$. It thus follows that $f(z)=0$ for all $z\in U$, and by dominated convergence, $f(is)=0$ for all $s\in\Bbb R$.