First thing to note is that $*r$ on the upper index is defined to be the convolutional power. This means that the affected function is convoluted with itself $r$ times.
The idea of the question is, that can the generalized binomial function be interpolated with the sinc functions? In other words, if $\binom{1}{x}= \operatorname{sinc}(x) + \operatorname{sinc}(x-1)$ is the interpolation of the binomial function. Is it possible to show that the generalization of the binomial function is the convolution of this interpolation?. This question comes from the observation that the binomial coeffiecients can be generated by convoluting a sequence $(1,1)$ by itself iteratively. I will provide my proof next and show that $$\big(\operatorname{sinc}(x) + \operatorname{sinc}(x-1) \big)^{*r} = \frac{\Gamma{(r+1)}}{\Gamma{(x+1)}\Gamma{(r-x+1)}}$$ indeed is a true proposition. This result underlines the fact, that the binomial function is the discrete sampling of the generalized version. Also, this formula implies, that the frequency domain of the generalized binomial function is bounded, because the fourier transform of the $sinc$-function is the rectangle and thus its domain is bounded. One interesting corollary is that, the binomial function can be represented in the following closed form $$\binom{n}{x} = \frac{\operatorname{sinc}{(x)}}{(1-x)^{\overline{n}}}n!$$ where $n\in\mathbb{N}$, $x\in \mathbb{C}$, and $(1-x)^{\overline{n}}$ represents the rising factorial. This can be proved by the induction. Another rather trivial corollary is that the generalized binomial function is a solution to the following differential equation $$x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \pi^2 xy=0$$ because $sinc(x-r+k)$ is a solution, thus the linear combination of these is also a solution, which implies that the general binomial function is a solution also. Some other applications could be possible, but these are the first that came to my mind. I hope someone finds this information usefull.
First of all, recall that the $\mathrm{sinc}$-function is defined to be $$\mathrm{sinc}(x) := \frac{\sin{(\pi x)}}{\pi x}$$ and is well-defined also at the origin. Next thing is to note that the $\mathrm{sinc}$-function is transitive in convolution.
$$ \mathrm{sinc}{(x-a)}*\mathrm{sinc}{(x-b)} = \mathcal{F}^{-1}\bigg(e^{-i2\pi\xi a}\chi\_{(-\frac{1}{2},\frac{1}{2})}(\xi)e^{-i2\pi\xi b}\chi_{(-\frac{1}{2},\frac{1}{2})}(\xi)\bigg)$$
$$ = \mathcal{F}^{-1}\bigg(e^{-i2\pi\xi (a+b)} \chi_{(-\frac{1}{2},\frac{1}{2})}(\xi)\bigg)$$
$$= \mathrm{sinc}{(x-(a+b))}$$
where $\chi_{(-\frac{1}{2},\frac{1}{2})}(\xi)$ is the characteristic, or the indicator, or the rectangle function depending whom to ask. This transitivity can be used to show that $$(\mathrm{sinc}{(x)} + \mathrm{sinc}{(x-1)})^{*r} = \sum_{k=0}^\infty \frac{(r)_k}{k!} \mathrm{sinc}{(x - r + k)}$$ hold for every $r\in\mathbb{R}$. The notation $ *r$ at the upper index means the $r^{th}$ convolution, and $(r)_k$ is defined to be the falling factorial.
The next step is to show that this convolution is identical to the generalized binomial function. The $\mathrm{sinc}$-function is related to the $\Gamma$-function trough the Euler reflection formula. $$\mathrm{sinc}{(x)} = \frac{1}{\Gamma{(x+1)}\Gamma{(1-x)}}$$ thus $$\frac{(r)_k}{k!} \mathrm{sinc}{(x - r + k)} = \frac{(r)_k}{k!} \frac{1}{\Gamma{(x - r + k+1)}\Gamma{(1-x + r - k)}}$$ The $\Gamma$-function has the property $\Gamma{(r+1)} = r\Gamma{(r)}$ which implies $$\frac{1}{\Gamma{(x-r+k+1)}} = \frac{1}{(x-r+1)^{\overline{k}}\Gamma{(x-r+1)}}$$ and $$\frac{1}{\Gamma{(1-x+r-k)}} = \frac{(r-x)_k}{\Gamma{(1-x+r)}}$$
At the following, the notation $(-r)^{\overline{k}}$ means the rising factorial. The falling and rising factorials are connected $(r)_k = (-1)^k(-r)^{\overline{k}}$, which can be seen when the terms from the falling factorial is turned negative one at a time. The number of terms is $k$, thus $(-1)^k$ is the sign of the modified factorial. $$\frac{(r)_k}{k!} = (-1)^k \frac{(-r)^{\overline{k}}}{k!}$$
$$(r-x)_k = (-1)^k (x-r)^{\overline{k}}$$
Combine everything above to get: $$\frac{(r)_k}{k!} \frac{1}{\Gamma{(x - r + k+1)}\Gamma{(1-x + r - k)}} = (-1)^k \frac{(-r)^{\overline{k}}}{k!} \frac{1}{(x-r+1)^{\overline{k}}\Gamma{(x-r+1)}} \frac{(-1)^k (x-r)^{\overline{k}}}{\Gamma{(1-x+r)}}$$
$$ = \frac{(-1)^{2k}}{\Gamma{(x-r+1)}\Gamma{(r-x+1)}}\frac{(-r)^{\overline{k}} (x-r)^{\overline{k}}}{(x-r+1)^{\overline{k}}}\frac{1}{k!}$$
Now take the sum over the index $k\in\mathbb{N}$. Note that $(-1)^{2k} = 1$ for all $k\in\mathbb{N}$. $$\sum_{k=0}^\infty \frac{(r)_k}{k!} \mathrm{sinc}{(x - r + k)} = \frac{1}{\Gamma{(x-r+1)}\Gamma{(r-x+1)}}\sum_{k=0}^\infty\frac{(-r)^{\overline{k}} (x-r)^{\overline{k}}}{(x-r+1)^{\overline{k}}}\frac{1}{k!}$$
The serie on the right hand side is the hypergeometric function $_2F_1(-r,x-r;x-r+1;1)$. This has a closed form in the terms of the $\Gamma$-function, which follows from the Gauss summation theorem. $$_2F_1(-r,x-r;x-r+1;1) = \frac{\Gamma{(x-r+1)}\Gamma{(r+1))}}{\Gamma{(x+1)}\Gamma{(1)}}$$ Substitute this result to the earlier equation, which had the hyperpolic function in it as a serie expansion. $$\sum_{k=0}^\infty \frac{(r)_k}{k!} \mathrm{sinc}{(x - r + k)} = \frac{1}{\Gamma{(x-r+1)}\Gamma{(r-x+1)}}\sum_{k=0}^\infty \frac{(-r)^{\overline{k}} (x-r)^{\overline{k}}}{(x-r+1)^{\overline{k}}}\frac{1}{k!}$$ $$= \frac{1}{\Gamma{(x-r+1)}\Gamma{(r-x+1)}} \frac{\Gamma{(x-r+1)}\Gamma{(r+1))}}{\Gamma{(x+1)}\Gamma{(1)}}$$
Reduce the equal terms from above and below to conclude that $$\sum_{k=0}^\infty \frac{(r)_k}{k!} \mathrm{sinc}{(x - r + k)} = \frac{\Gamma{(r+1)}}{\Gamma{(x+1)}\Gamma{(r-x+1)}}$$
Because the left and right hand side of the equation is well defined on the whole set of the complex numbers, this equation also holds when $r,x\in\mathbb{C}$.