Your questions are a part of what Deninger has been writing about for 20 years. He's proposed a point of view that sort of explains a lot of things about zeta functions. It's important to say that this explanation is more in a theoretical physics way than in a mathematical way, in that, as I understand it, he's predicted lots of new things which he and other people have then gone on to prove using actual mathematics. I guess it's kind of like the yoga surrounding the Weil conjectures before Dwork and Grothendieck made actual cohomology theories that had a chance to do the job (and eventually did). It's pretty clear to me that he's put his finger on something, but we just don't know what yet.
Let me try to say a few things. But I should also say that I never worried too much about the details, because the details he has are about a made up picture, not the real thing. (If he had the real thing, everyone would be out of a job.) So my understanding of the actual mathematics in his papers is pretty limited.
Question 1: He gives some evidence that Euler factors at both finite and infinite places should be seen as zeta-regularized characteristic polynomials. For the usual Gamma function, see (2.1) in [1]. For the Gamma factors of general motives, see (4.1) in [1]. For the Euler factors at the finite places, see (2.3)-(2.7) in [2]. He gives a description that works simultaneously at the finite and infinite places in (0.1) of [2]. Beware that some of this is based on an artificial cohomology theory that is designed to make things uniform over the finite and infinite places. (Indeed, at the risk of speaking for him, probably the whole point was to see what such a uniform cohomology theory would look like, so maybe one day we'll be able to find the real thing.)
Question 2: He expects his cohomology theory to have a Poincare duality which is "compatible with respect to the functional equation". See the remarks and references in [3] between propositions 3.1 and 3.2.
I'd recommend having a look at [3]. It's mainly expository. Also, I remember [4] being a good exposition, but I don't have it in front of me now, so I can't say much. He also reviews things in section 2 of his recent Archive paper [5].
[1] "On the Gamma-factors attached to motives", Invent. Math. 104, pp 245-261
[2] "Local L-factors of motives and regularized determinants", Invent. Math. 107, pp 135-150
[3] "Some analogies between number theory and dynamical systems on foliated spaces", Proceedings of the ICM, Vol. I (Berlin, 1998), pp 163-186
[4] "Evidence for a cohomological approach to analytic number theory", First ECM, Vol. I (Paris, 1992), pp 491-510
[5] "The Hilbert-Polya strategy and height pairings", arxiv.org
$\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}\newcommand\Z{\mathbb Z}$The answer is yes. Indeed, the conditions $f\colon A\to\R$ and $f(x+1)=xf(x)$ for all $x\in A$ imply that $x+1\in A$ for any $x\in A$ and hence, by induction, $x+k\in A$ for all natural $k$. We also have $(0,\infty)\subseteq A\subseteq\R$ and $f(x)=\Ga(x)$ for all $x>0$.
Now take any $y\in A\cap(-\infty,0]\setminus \{0,-1,-2,\dots,\}$. It remains to show that $f(y)=\Ga(y)$. Take any natural $n$ such that $y+n>0$. Then $\Ga(y+n)\ne0$ and
\begin{equation}
f(y)\prod_{k=0}^{n-1}(y+k)=f(y+n)=\Ga(y+n)=\Ga(y)\prod_{k=0}^{n-1}(y+k). \tag{1}
\end{equation}
So, $\prod_{k=0}^{n-1}(y+k)\ne0$ and $f(y)=\Ga(y)$, as desired. (The conclusion $\prod_{k=0}^{n-1}(y+k)\ne0$ also follows because $y$ is not an integer and hence $y+k\notin\mathbb Z$ for any integer $k$.)
Best Answer
I assume you have already found this page? It seems that, from the definition $$ H(x) = \frac{\Psi(1-x/2)-\Psi(1/2-x/2)}{2\Gamma(1-x)} $$ the various properties you are interested in should be straightforward to prove.