Let $f_p = 1_{\mathbb{Z}p}$ for $p$ finite and $f_\infty(x) = e^{-\pi x^2}$ for $p$ at infinity, and $f = \otimes_v f_v$ where the product is over the places of $\mathbb{Q}$.
Then the zeta integral of $f$ with respect to the character $||^s$ gives the continuation of the zeta function: $$Z(f, ||^s) = \pi^{-s/2}\Gamma(s/2)\zeta(s).$$
Is there a way to recover the factor $s(s-1)$ that appears in the xi function? What about in the number field case where one gets an additional factor of $|\Delta_K|^{s/2}$ (the norm of the discriminant)?
I can understand the impulse to obtain the most-symmetrical integral representation... and I recall trying to make it happen, myself. :)
But, assuming that we don't want the test function to depend on $s$, I think (having tried both concretely and abstractly) there's no natural candidate, and perhaps none at all, so obtain the $s$ and $s-1$ factors.
Likewise, to get the conductor, the natural method (to my perception) is to just take characteristic functions of local integers everywhere, and then symmetrize at the end. For some rings of algebraic integers, it is possible to symmetrize this by choice of test function, but not for all...
Perhaps illustrating the obstacle(s) more definitively, consider the case of an $L$-function with real character, so that its functional equation relates it to itself (with the power of conductor...). But there is also that pesky epsilon factor in the functional equation, which already over $\mathbb Q$ is the fairly tricky argument-of-Gauss-sum (after normalizing-away the powers of conductor).
Even the more-fancy way to talk about local $L$-factors as greatest common divisors doesn't get rid of powers of the conductor in an absolutely convincing fashion. :)