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For $a>0$ let us define $$H^2(-a,a)=f mbox<a$: sup_yin [-a,a]int_mathbbR.$$ For $fin H^2(-a,a)$, define $|f|=sup_yin (-a,a)int_mathbbR|f(x+iy)|^2,dx<infty$. We note that the function $e^-z^2in H^2(-a,a)$ for any $a>0$. Can we have that
$operatornamespan(e^-(z-b)^2: binmathbbR)$ is dense in $H^2(-a,a)$ with respect to the norm $|cdot|$?

I'm not sure. (Edit: actually I believe that yes, those functions are dense. See below) Some comments on this $H^2(-a,a)$ thing that may be helpful:

First, it's not at all clear tat the norm you define is, or is equivalent to, a Hilbert-space norm. But in fact if $fin H^2(-a,a)$ then the boundary values $f(xpm ia)$ exist almost everywhere, and the norm above is equivalent to the $L^2$ norm of the boundary values.

Second, in fact $fin H^2(-a,a)$ if and only if there exists $Fin L^2(Bbb R)$ with $$int e^|F(xi)|^2,dxi<infty$$and $$F(z)=int e^izxiF(xi),dxi;$$in fact $||f||$ is equivalent to $$left(int e^|F(xi)|^2,dxiright)^1/2.$$So your question reduces to the question of whether the span of the functions $e^ibxie^-xi^2$ is dense in $H=L^2left(e^dxiright)$.

I think it's clear that this is so. Say $F_b(xi)=e^ibxie^-xi^2$. Say $Fin H$ and $Fperp F_b$ for every $b$. If you let $G(xi)=e^e^-xi^2F(xi)$ then $Gin L^1(Bbb R)$ and $hat G=0$; hence $G=0$, so $F=0$.

About those preliminary results, the second in particular: I don't have a good reference. (i) Many years ago I heard Rudin say this had beenn proved by Bochner; I don't have a reference for that, to Rudin or to Bochner. (ii) I suspect this is a special case of results on "tube domains" in Stein&Weiss "Harmonic Analysis...".("Fourier Analysis..."?)

(iii) If you want to prove it yourself this might be a way to get started: Say $f_y(x)=f(x+iy)$. Let $f_pm a$ be a weak limit point of $f_y$ as $ytopm a$. Then a simple limiting argument shows that $$f(z)=frac12pi ileft(intfracf_-a(t)z-(t-ia),dt-intfracf_a(t)z-(t+ia),dtright).$$

Hint for showing the integral over the segments constituting the ends of the rectangles tends to $0$: If $0<delta<a-|y|$ then
$$beginalign|f(x+iy)|^2&=frac1pidelta^2left|iint_s^2+t^2<delta^2f^2(x+iy+s+it),dsdtright|\&lefrac1pidelta^2int_-delta^deltaint_-delta^delta|f(x+iy+s+it)|^2,dsdtendalign$$