Computing the Fourier transform of $H_k(x)e^{-x^2/2}$, where $H_k$ is the Hermite polynomial.

Yes, once you have the equation $$ \sum_{n}\mathscr{F}(e^{-x^2/2}H_n(x))\frac{t^n}{n!} = \sum_{n}\sqrt{2\pi}(-i)^n H_n(\xi)e^{-\xi^2/2}\frac{t^n}{n!}, $$ then you can fix $\xi$, and view this as a power series equation that holds for all $t$. Therefore, for this fixed $\xi$, the power series coefficients must be identical, which leads to $$ \mathscr{F}(e^{-x^2}H_n(x))=\sqrt{2\pi}(-i)^n H_n(\xi) $$ The left side implicitly depends on $\xi$, because it is the transform variable. The transform is continuous in $\xi$ because of the exponentially decaying nature of the function being transformed. So $$ \mathscr{F}(e^{-x^2}H_n(x))(\xi)=\sqrt{2\pi}(-i)^n H_n(\xi),\;\;\xi\in\mathbb{R}. $$ Omitting the arguments of the functions, these functions are equal: $$ \mathscr{F}(e^{-x^2}H_n(x)) = \sqrt{2\pi}(-i)^n H_n $$

Once you know that the series in question converges absolutely in $L^2$ then the identification becomes automatic. Writing $\psi_k(x) = \frac{1}{k!} H_k(x) e^{-x^2/2}$, $k\geq 0$ then each $\psi_k$ is an element in the Hilbert space $L^2({\Bbb R})$ having norm $\frac{2^{k/2} } {\sqrt{k!}} \pi^{1/4}$. One has $\lim_k \|\psi_k\|_{L^2}^{1/k} = 0 $, so the series $$ \Psi_t(x) = \lim_{N\rightarrow \infty} \sum_{k=0}^{N-1} t^k \psi_k(x)$$ is norm convergent in $L^2$ for all complex $t$. $\Psi_t$ is therefore an analytic $L^2$ valued function of $t\in {\Bbb C}$. Coefficients in an analytic series are uniquely determined by the function and may be recovered e.g. by a Cauchy integral (which works fine in Hilbert or Banach spaces). You may also recover them simply from looking at successive derivatives at $0$, as you mention. Since Fourier transformation (denoted by a 'hat') preserves the $L^2$ norm we automatically get: $$ \widehat{\Psi}_t(\xi) = \lim_{N\rightarrow \infty} \sum_{k=0}^{N-1} t^k \widehat{\psi}_k(\xi),$$ again an analytic $L^2$ valued function. An issue is how to calculate the two sums (as $L^2$ limits) but as you say, in your book they prove that $\Psi_t(x)$ (in $L^2$) is identical to $$ \Phi_t= \exp \left( 2xt-t^2-x^2/2 \right) $$ This then resolves the problem since as elements of $L^2$: $$\widehat{\Psi}_t(\xi)= \widehat{\Phi}_t(\xi) = \exp \left( -2it\xi+t^2-\xi^2/2 \right) = \Phi_{-it}(\xi) =\Psi_{-it}(\xi)$$ Identifying the analytic expansions for $\widehat{\Psi}_t(\xi)=\Psi_{-it}(\xi)$ you must have: $\widehat{\psi_k}(\xi) = (-i)^k \psi_k(\xi)$ in $L^2$ (where possibly a factor of $\sqrt{2\pi}$ should be included). As both are actually smooth functions, you get the wanted identity.

Returning to the limit in $L^2$, let me provide a relatively simple proof: The defining equation, $H_k(x)=(-1)^k e^{x^2} \partial_x^k e^{-x^2}$ shows that $\psi_k=\frac{1}{k!}H_k(x) e^{-x^2/2}$ itself may be expressed through $\Phi_t$ as a Cauchy integral as follows: $$ \psi_k(x)= (-1)^k e^{x^2/2} \oint_{\partial B(x,R)} \frac{e^{-z^2}}{(z-x)^{k+1}} \frac{dz}{2\pi i} = \oint_{\partial B(0,R)} \frac{e^{-x^2/2+2xu-u^2}}{u^{k+1}} \frac{du}{2\pi i}$$ where $R>0$ is arbitrary. From this we get the bound $|\psi_k(x)| \leq R^{-(k+1)} e^{-x^2/2+2|x|R +R^2}$ which is clearly square integrable (in $x$) so that $\psi_k$ is an element in $L^2$.

Now fix a value of $t$ and let $R>|t|$. Then by summing a geometric series: $$ \sum_{k=0}^{N-1} t^k \psi_k(x) = \oint_{\partial B(0,R)} (1 - (t/u)^N)\frac{e^{-x^2/2+2xu-u^2}}{u-t} \frac{du}{2\pi i}= \Phi_t(x) + {\cal R}^{N}(t,x)$$ where we have the following bound for the remainder: $$ |{\cal R}^{N}(t,x)| \leq \frac{(|t|/R)^N}{R-|t|} e^{-x^2/2+2R|x|+R^2}.$$ This is indeed a function in $L^2({\Bbb R})$ and goes to zero in norm when $N\rightarrow \infty$, thus justifying the above identifications.