To your first question, yes, it does make sense to define $P(n,r)=\prod_{i=0}^{r-1}(n-i)$, thus extending domain of $n$ to all of $\mathbb C$ (or really any commutative ring with identity). Then $P(n,r)$ is also called the falling factorial, sometimes denoted $n^{\underline r}\,$.
Just like $C(n,k)$ counts sets of size $k$ for $n>0$, there is a nice combinatorial interpretation $|C(-n,k)|$; it counts multisets of size $k$. There is a similar duality for $P(n,k)$. Whereas $P(n,k)$ counts ordered lists without repetition, $|P(-n,k)|$ counts the number of ways to place $k$ distinct flags on $n$ distinct flagpoles, where the order of the stack of flags on each pole matters. Note that $|P(-n,k)|=n(n+1)\cdots (n+k-1)$ is the rising factorial, denoted $n^{\overline k}$. This is sort of like an ordered version of a multiset. You can think of the flagpoles as the objects being chosen, and putting a flag on it an object represents choosing it. You can put multiple flags on a pole, so you can choose an object multiple times. Since the flags are distinct, order matters.
Finally, the reason that Wolfram Alpha cannot compute $P(n,r)$ when $n$ is a negative integer is because $(-n)!$ is undefined. Specifically, the Gamma function has a simple pole at each nonnegative integer. However, we can still calculate $P(n,r)$ using $n!/(n-r)!=\Gamma(n+1)/\Gamma(n-r+1)$. Since we have a simple pole in the numerator and denominator, they effectively cancel. The ratio now has a removable singularity at each nonnegative integer, and we can define $P(n,r)$ as the limit for $n$ in the neighborhood of the negative integer. In this case, we recover the expected $P(n,r)=\prod_{i=0}^{r-1}(n-i)$ result.
Not an answer, but too long for a comment: I wouldn't expect a closed form.
If $r$ is a nonnegative integer, then (as I show in this answer) $$\binom{r}{t}=\frac{\sin\pi t}{\pi}\sum_{k=0}^r\binom{r}{k}\frac{(-1)^k}{t-k};$$ thus, even at $z=1$, $\int_0^\infty\binom{r}{t}\,dt$ is a combination of values of the sine integral.
Let's replace $z$ with $e^{-z}$ (now we assume $\Re z>0$). So, if $r$ is a nonnegative integer, $\int_0^\infty\binom{r}{t}e^{-zt}\,dt$ expresses in terms of $\int_0^x\frac{\sin\pi t}{\pi t}e^{zt}\,dt$ at integer values of $x\leqslant r$ (and some elementary extra things). Clearly, the situation when $r$ is not an integer is even more complicated. Anyway, alternative integral representations come from the ones for $\binom{r}{t}$, that is, for the "reciprocal beta".
Say, if $\Re(a+b)>-1$ and $0<c<1$, we have $$\int_{c-i\infty}^{c+i\infty}w^{-a-1}(1-w)^{-b-1}\,dw=\frac{2\pi i\,\Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)},\tag{I}\label{intrep}$$ which gives a representation for $\binom{r}{t}$, $\Re r>-1$ if we put $a=t$ and $b=r-t$; this results in $$\int_0^\infty\binom{r}{t}e^{-zt}\,dt=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}w^{-1}(1-w)^{-r-1}\big(z+\log w-\log(1-w)\big)^{-1}dw$$ for $1/2<c<1$ (at least). If we deform the path of integration to encircle any of the branch cuts of the integrand, we obtain real alternative integral representations. The same can be done for $\Re r\leqslant-1$.
Update. The integral representation \eqref{intrep} can be used to show that
$$\int_{-\infty}^\infty\binom{s}{x}\,dx=2^s\qquad(\Re s>-1)$$ (stated in this answer). We take $c=1/2$ and substitute $w=\displaystyle\frac12\left(1-i\tan\frac\phi2\right)$: $$\binom{s}{x}=\frac{2^s}{2\pi}\int_{-\pi}^\pi\left(1+i\tan\frac\phi2\right)^{-s}e^{ix\phi}\,d\phi,$$ which, integrated over $x\in(-y,y)$, after substituting $\phi=t/y$, gives $$\int_{-y}^y\binom{s}{x}\,dx=\frac{2^s}{\pi}\int_{-y\pi}^{y\pi}\left(1+i\tan\frac{t}{2y}\right)^{-s}\frac{\sin t}{t}\,dt.$$ Now take $y\to\infty$ (DCT is clearly applicable here).
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {m - 1/2 \choose m} & = {\pars{m - 1/2}! \over m!\pars{-1/2}!}= {\Gamma\pars{m + 1/2} \over m!\,\Gamma\pars{1/2}} \\[5mm] & = {1 \over m!\,\root{\pi}}\,\ \overbrace{{\root{2\pi}2^{1/2 - 2m}\,\Gamma\pars{2m} \over \Gamma\pars{m}}} ^{\ds{\color{#f00}{\large\S}}\,,\ \Gamma\pars{m + 1/2}}\,,\quad \pars{~\Gamma\pars{1 \over 2} = \root{\pi}~} \\[5mm] & = {1 \over 2^{2m - 1}}\,{\pars{2m - 1}! \over m!\pars{m - 1}!} = {1 \over 2^{2m - 1}}\,{\pars{2m}!/\pars{2m} \over m!\pars{m!/m}} \\[5mm] & = {1 \over 2^{2m}}\,{\pars{2m}! \over m!\, m!} = \color{#f00}{{1 \over 2^{2m}}{2m \choose m}} \end{align} $\ds{\color{#f00}{\large\S}:\ \Gamma\!-\!Duplication\ Formula}$. See $\ds{\mathbf{6.1.18}}$ in Abramowitz & Stegun Table.
$$ {2m \choose m} = 2^{2m}{m - 1/2 \choose m} = 2^{2m}\bracks{{-1/2 \choose m}\pars{-1}^{m}} = {-1/2 \choose m}\pars{-4}^{m} = {-1/2 \choose -1/2 - m}\pars{-4}^{m} $$