Not an answer, but maybe helpful.

These questions can be much harder than they look. There is a simple-looking, explicit set of functions $\{f_\alpha \} \subset L^2(0,1; dx/x)$ (see Operators, Functions, and Systems: an Easy Reading: Volume 1 by Nikolai K. Nikolski) with the property: the Riemann Hypothesis is *equivalent* to the density of the span of $\{f_\alpha \}$.

Of course it doesn't mean your problem is so difficult; but it's certainly interesting and non-trivial (I think). The necessary results should be known and exist somewhere [*UPDATE: I've changed my mind, maybe not!!*] I haven't found anything yet; but the general problem I describe below is certainly very "natural" and I'm sure I'm not the first to think of it, so if it isn't known then it's an interesting open problem; I would be very interested to know the solution!

The Laplace transform gives (a multiple of) a unitary operator from $L^2(0, \infty)$ onto the Hardy space $H^2( \{ \mathrm{Re}(z)>0 \})$ (depending on your normalisations).

The Laplace transform of $x^{n} e^{-x/n}$ is, up to a constant, $(z+1/n)^{-(n+1)}$.

So you're asking whether the orthogonal complement of these functions is zero in $H^2$. The scalar product of $F(z)$ with $(z+\overline{\lambda})^{-k-1}$ is, up to a constant, the derivative $F^{(k)}(\lambda)$.

Thus (assuming my algebra is correct), your question is equivalent to asking whether there is any non--trivial $F \in H^2$ satisfying

$$
F^{(n)}(1/n) = 0, \qquad n=1,2,3,\ldots
$$

If you just wanted some non-trivial analytic function $F$ on the half-plane $ \{ x+iy : x>0 \}$ to satisfy this, it's possible (I think, if I remember correctly!) - at each point of a countable set without limit point in the domain, we can prescribe values of finitely many derivatives. The extra condition $F \in H^2$ is the difficult part.

More generally we have:

**Problem** classify all sequences $(z_n)$, $(k_n)$ such that we have a uniqueness result:
$$
F \in H^2, \quad F^{(k_n)}(z_n) =0 \quad (n=1,2,3,\ldots) \qquad \Rightarrow \qquad F \equiv 0.
$$

Of course there might be a special trick for your particular case $z_n = 1/n$, $k_n = n$.

Special cases are well-known. For example: there is no non-trivial $G \in H^2$ satisfying $G(z_n)=0$ *if and only if* $\sum_n \frac{\mathrm{Re}(z_n)}{|1+z_n|^2} = +\infty$ (the Blaschke condition). e.g. consider $z_n$ converging to zero, or out to infinity; if it does this so quickly that the sum is finite, then the set $\{ z_n \}$ is sparse enough to allow non-trivial $G$. Thus the classification is known if $k_n = 0$ for all $n$.