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

Using the notation $$g(x)risingdotseq G(s) quadtext{if}quad G(s)=mathcal L(g(x)),$$ one can write $$M_{large s}(f(t)) = intlimits_0^1 t^{large s} f(t),mathrm dt = intlimits_0^infty e^{large-sx}e^{large-x}f(e^{large-x}),mathrm dx =e^{large-s^2},$$ $$g(x) = e^{large-x}fleft(e^{large-x}right)risingdotseq e^{large-s^2},tag1$$ $$dfrac{sqrtpi}2operatorname{erf} s = intlimits_0^{large s}M_{large s}(f(t)),mathrm dt = intlimits_0^1 dfrac{t^{large s}}{ln t}f(t),mathrm dt = intlimits_0^infty e^{large-sx}dfrac1x e^{large-x}f(e^{large-x}),mathrm dx,$$ $$dfrac1x g(x)risingdotseq dfrac{sqrtpi}2operatorname{erf} s.tag2$$ Since $$e^{large-frac14x^2}risingdotseqsqrtpi e^{large s^2}operatorname{erf}s,tag3$$ then $$e^{large-frac14x^2}*g(x) = dfrac2x g(x),$$ $$2g(x) = xintlimits_0^{large x} e^{large-frac14(x-t)^2}, g(t),mathrm dt = 2intlimits_0^{large x} g(t) left(e^{large-frac14(x-t)^2}right)'_t,,mathrm dt + intlimits_0^{large x} e^{large-frac14(x-t)^2}, tg(t),mathrm dt \ overset{IBP}{=!=!=}, 2 g(t) left(e^{large-frac14(x-t)^2}right)bigg|_0^{large x} - 2intlimits_0^{large x}e^{large-frac14(x-t)^2}g'(t),,mathrm dt + intlimits_0^{large x} e^{large-frac14(x-t)^2}, tg(t),mathrm dt,$$ $$intlimits_0^{large x} e^{large-frac14(x-t)^2}, (tg(t)-2g'(t)),mathrm dt = g(0)e^{large-frac14x^2},$$ $$intlimits_0^{large x} e^{largefrac12 xt}e^{large-frac14t^2}, (tg(t)-2g'(t)),mathrm dt = g(0).tag4$$ $RHS(4)=mathrm{constant}(x),$ so the solution is defined via ODE task $$g'(x) = frac t2 g(x),quad g(0)= 0,tag5$$ without reguar non-zero solutions.

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 $$lim_{ntoinfty} r^n M_n(mu) = 0 tag{*} $$ 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_{Ntoinfty} S_N(r) = 0 $ for any $r > 0$. Moreover, by the Fubini's Theorem and the Dominated Convergence Theorem,

begin{align*} 0 &= lim_{Ntoinfty} S_N(r)\ &= lim_{Ntoinfty} 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_{Ntoinfty} int_{[0,1]} left( 1 - e^{-(t/r)^N} right) , mu(mathrm{d}t) \ &= int_{[0,1]} lim_{Ntoinfty} 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 $0$. $square$