Can Laplace's transformation be equal to a Gaussian for any integer?

We prove a more general claim:

Claim. Let $\mu$ be a signed finite Borel measure on $[0, 1]$ and write $M_n(\mu) = \int_{[0,1]} t^n \, \mu(\mathrm{d}t)$. Suppose $$\lim_{n\to\infty} r^n M_n(\mu) = 0 \tag{*} $$ holds for any $r > 0$. Then $\mu = c \delta_0$ for some constant $c$.

Note that OP's case corresponds to a signed measure of the form $\mu(\mathrm{d}t) = f(t) \, \mathrm{d}t$. Then the claim tells that there exists no such $\mu$ satisfying $M_n(\mu) = e^{-n^2}$ eventually. Indeed, any such $\mu$ would satisfy $\text{(*)}$, and then the claim leads to a contradiction that $M_n(\mu) = 0$ for all $n \geq 1$.

Proof of Claim. Assume that $\text{(*)}$ holds. For any $r > 0$ and $N \in \mathbb{N}_1$, we define

$$ S_N(r) := \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} r^{-Nk} M_{Nk}(\mu). $$

Then from the bound

$$\left| S_N(r) \right| \leq \sum_{k=1}^{\infty} \frac{1}{k!} r^{-Nk}\left| M_{Nk}(\mu)\right| \leq e \sup_{n \geq N} \left( r^{-n}\left| M_n(\mu) \right| \right), $$

we have $ \lim_{N\to\infty} S_N(r) = 0 $ for any $r > 0$. Moreover, by the Fubini's Theorem and the Dominated Convergence Theorem,

\begin{align*} 0 &= \lim_{N\to\infty} S_N(r)\\ &= \lim_{N\to\infty} \int_{[0,1]} \left( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} (t/r)^{Nk} \right) \, \mu(\mathrm{d}t) \tag{$\because$ Fubini} \\ &= \lim_{N\to\infty} \int_{[0,1]} \left( 1 - e^{-(t/r)^N} \right) \, \mu(\mathrm{d}t) \\ &= \int_{[0,1]} \lim_{N\to\infty} \left( 1 - e^{-(t/r)^N} \right) \, \mu(\mathrm{d}t) \tag{$\because$ DCT} \\ &= \int_{[0,1]} \left( \mathbf{1}_{\{t > r\}} + (1-e^{-1})\mathbf{1}_{\{t=r\}} \right) \, \mu(\mathrm{d}t) \\ &= \mu([r,1])-e^{-1}\mu(\{r\}). \end{align*}

(When $r > 1$, we regard $[r, 1] = \varnothing$.) Consequently,

$$ \mu([r, 1]) = 0 $$

holds, initially when $r$ is not an atom of $\mu$, and then for all $r > 0$ by the limiting argument. Therefore $\mu$ must be concentrated at $0$. $\square$