Multiple answers and comments have already pointed out that the conceptual role of $\pi^{-s/2}\Gamma(s/2)$ comes from the viewpoint of Tate's thesis, which for $\text{Re}(s) > 1$ creates this function as $\int_{\mathbf R^\times} e^{-\pi x^2}|x|^s\,dx/|x|$, an integral over the multiplicative group $\mathbf R^\times$ of the function $e^{-\pi x^2}$ that is self-dual for the Fourier transform on the additive group $\mathbf R$ relative to the self-duality $\langle x,y\rangle = e^{2\pi ixy}$ or $\langle x,y\rangle = e^{-2\pi ixy}$ on $\mathbf R$. (If we use another self-duality of $\mathbf R$ then $e^{-ax^2}$ would be self-dual for some $a \not= \pi$ instead.)

It's also been said elsewhere on this page that there are *many* self-dual Schwartz functions on $\mathbf R$, or more specifically *many even* self-dual Schwartz functions on $\mathbf R$: for Schwartz $f$ on $\mathbf R$ and $\text{Re}(s) > 0$, we have $\int_{\mathbf R^\times} f(x)|x|^s\,dx/|x| = \int_{0}^\infty (f(x) + f(-x))x^s\,dx/x$ and this is $0$ when $f$ is odd, so we may as well assume $f$ is even since $f(x) + f(-x)$ is even anyway and we want to avoid the silly equation $0=0$ even if it is a valid equation.

For arbitrary Schwartz $f$ on $\mathbf R$, set $\Gamma_f(s) = \int_{0}^\infty f(x)x^s\,dx/x$, which is a mild modification of the function $\Gamma(f,s)$ in Paul Garrett's answer (his $\Gamma(f,s)$ is my $\Gamma_{f(x)+f(-x)}(s)$ by a formula I wrote in the previous paragraph). This function converges absolutely and is analytic for $\text{Re}(s) > 0$, and it extends meromorphically to $\mathbf C$ by repeated integration by parts (the same way the $\Gamma$-function can be extended to $\mathbf C$ from its integral definition for $\text{Re}(s) > 0$), and Tate's thesis shows there is a general functional equation $\Gamma_f(s)\zeta(s) = \Gamma_{\hat{f}}(1-s)\zeta(1-s)$ where $\hat{f}$ is the Fourier transform of $f$ (for the self-duality on $\mathbf R$ given by $\langle x,y\rangle = e^{-2\pi ixy}$), so if $f$ is self-dual then we get $$\Gamma_f(s)\zeta(s) = \Gamma_{f}(1-s)\zeta(1-s),$$ a very nice functional equation indeed, especially if we use even $f$ to avoid $0 = 0$. 

All of what I wrote so far has appeared explicitly or implicitly in some of the other comments or answers. Since there are *many* self-dual even Schwartz functions $f$ on $\mathbf R$, what is it about the choice $f(x) = e^{-\pi x^2}$, leading to $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$ (an extra $1/2$ on both sides of the functional equation can be cancelled) that is so nice? I have not seen the following property pointed out yet: with this choice of $f$ and familiarity with the $\Gamma$-function *we know* $\Gamma_f(s) \not= 0$ *for* $\text{Re}(s) > 1$ (in fact for $\text{Re}(s) > 0$), so therefore $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) > 1$ from $\zeta(s)$ being nonvanishing there, and then by the functional equation $\Gamma_f(s)\zeta(s) \not= 0$ for $\text{Re}(s) < 0$, which means all zeros of $\Gamma_f(s)\zeta(s)$ have $0 \leq \text{Re}(s) \leq 1$.  If you want to use a totally random even Schwartz function for $f$ in order to complete the Riemann zeta-function as in Tate's thesis, you get the nice-looking nontrivial functional equation displayed above, but how are you going to use $\Gamma_f(s)\zeta(s)$ to analyze the location of zeros of $\zeta(s)$ (including discovering its  trivial zeros, whether or not you consider those important) if you do not know where $\Gamma_f(s)$ has its zeros and poles? 

So although there are many even Schwartz functions $f$ on $\mathbf R$ besides $e^{-\pi x^2}$ that you could use to get a nice functional equation by multiplying $\zeta(s)$ by $\Gamma_f(s)$, the reason that the choice $f(x) = e^{-\pi x^2}$ is so convenient is that we actually *know* the zeros and poles of $\Gamma_f(s) = (1/2)\pi^{-s/2}\Gamma(s/2)$: it has no zeros in $\mathbf C$ and it has simple poles at $0, -2, -4, \ldots$. For even self-dual Schwartz $f$ on $\mathbf R$ that are not simple modifications of $e^{-\pi x^2}$, how feasible is it to determine whether or not $\Gamma_f(s) \not= 0$ for $\text{Re}(s) > 1$ (or $\text{Re}(s) > 0$)?  The method of meromorphically continuing $\Gamma_f(s)$ from the half-plane $\text{Re}(s) > 0$ where it is analytic to all of $\mathbf C$ shows that its only possible poles are at $0, -1, -2, -3, \ldots$ with orders *at most* $1$ and the residue at $s = -n$ is $(-1/n!)\int_0^\infty f^{(n+1)}(x)\,dx$, which by the Fundamental Theorem of Calculus is $(-1/n!)(f^{(n)}(\infty) - f^{(n)}(0)) = f^{(n)}(0)/n!$.
Therefore you could determine the poles of $\Gamma_f$ by seeing when $f^{(n)}(0)$ is 0 and not 0, but how are you going to determine where the *zeros* of $\Gamma_f$ are or that there are no zeros? (EDIT: for even $f$, its odd-order derivatives vanish at $0$, so the residue at $-n$ vanishes when $n$ is odd, which means the poles of $\Gamma_f(s)$ can only be at $n = 0, -2, -4, -6, \ldots$. Those are *all* simple poles of $\pi^{-s/2}\Gamma(s/2)$, which has no zeros, so $\Gamma_f(s)/(\pi^{-s/2}\Gamma(s/2)$ is an entire function, which makes $\pi^{-s/2}\Gamma(s/2)$ a "holomorphic gcd" of all $\Gamma_f(s)$ for even Schwartz functions $f$ on $\mathbf R$. This addresses a comment below by Venkataramana.)

Example: the function $f(x) = 1/(e^{\pi x} + e^{-\pi x})$ is an even self-dual Schwartz function on $\mathbf R$. Can someone determine in a self-contained way (i.e., not using $\zeta(s)$) where $\Gamma_f(s)$ has its zeros on $\mathbf C$, or determine if it has no zeros? 

Edit: Ignoring the wacky example just above, in some comments below I work out an example with $f(x)$ being a 4th degree Hermite polynomial times a Gaussian and find that $\Gamma_f(s)$ has two zeros with positive real part, at $s = (1\pm \sqrt{-2})/2$.