Gamma-free definition of binomial coefficients
Yes, there is a Bohr-Mollerup (or Artin-Bohr-Mollerup?) criterion for $n \choose x$. It's probably known, but seems easier to write up anew than to find in the literature.
Proposition: The function $n\choose x$ of an integer $n \geq 0$ and a real $x$ is characterized by the initial value ${0 \choose 0} = 1$ and the recurrences $$ (x+1) {n \choose x+1} = (n-x) {n \choose x}, \quad\ {n \choose x} + {n \choose x+1} = {n+1 \choose x+1} $$ (which imply that $n \choose x$ agrees with the usual values when $x \in \bf Z$), together with the condition that $n \choose x$ is a logarithmically convex function of $x \in (0,n)$ for each $n$.
(We do not need to impose the symmetry condition ${n \choose x} = {n \choose n-x}$, but it follows automatically from uniqueness.)
Proof sketch: We can show that the trigonometric $n \choose x$ satisfies this convexity condition using the product formula for the sine: each factor $(x+k)(x-n-k)$ (with $k=0,1,2,\ldots$) is logarithmically convex (downwards) on $0 \leq x \leq n$, so the same is true of their product.
The converse can be proved a la Bohr-Mollerup. Any function of $n$ and $x$ satisfying the recurrences is of the form $f(x) {n \choose x}$ for some $1$-periodic function $f$. But for large $n$ and $x$ near $n/2$ the logarithm of $n \choose x$, though convex, is nearly flat; e.g. $$ \log {2m \choose m-1} - 2 \log {2m \choose m} + \log {2m \choose m+1} = O(1/m). $$ Thus if $f$ were nonconstant then for large $n$ the fluctuations would make $f(x) {n \choose x}$ nonconvex. Hence $f$ is constant, and then the initial value ${0 \choose 0} = 1$ makes $f(x)=1$ identically. $\Box$
Let $\ \mu\ $ be a measure in $\ \mathbb C.\ $ Then define $\ \mathbb P_\mu(z)$:
$$ \mathbb P_\mu(z)\ :=\ e^{ \int_\mu {\log(z-t)\,dt} } $$
We may conveniently define also the respective Newton coefficient:
$$ \binom z{\mu,n}\ :=\ \frac{\mathbb P_\mu(z)}{\mathbb P_\mu(n)} $$
when $ \mathbb P_\mu(n) \ne 0\ $ and is well-defined (for arbitrary $\ n\in\mathbb C$) -- this coefficient is more like a generalized polynomial.
EXAMPLE Let $\ \mu\ $ be an atomic measure where $\ \mu(\{k\}) := 1\ $ for every $\ k=0\ \ldots\ n\!-\!1\ $ (this time $\ n\ $ is a natural number), and there are no other atoms. Then $\ \binom z{\mu,n}\ $ is equal to a respective basic integer polynomial or Newton polynomial (coefficient) $\ \binom zn.$
This my definition is just a draft of a note.
EDIT Let me add another definition along similar lines (a measure). My note is a draft since--for instance--I am not discussing here the question of the set of arguments of the introduced functions; also, for the second definition below, one should also discuss the question of the measure staying in some sense away from $\ 0\in\mathbb C\ $ (e.g. let $\ \mu(B(0; r)) = 0\ $ for certain $\ r>0\ $ (but much less can and should be assumed).
Let $\ \mu\ $ be a measure in $\ \mathbb C.\ $ Then define $\ \mathbf p_\mu(z)$:
$$ \mathbf p_\mu(z)\ :=\ e^{ \int_\mu {\log(\frac zt-1)\,dt} } $$
We may conveniently define also the respective Newton coefficient:
$$ \binom z{\mu;n}\ :=\ \frac{\mathbf p_\mu(z)}{\mathbf p_\mu(n)} $$
when $ \mathbf p_\mu(n) \ne 0\ $ and well-defined (for arbitrary $\ n\in\mathbb C$).
WARNING This time the Newton coefficient notation features ";" instead of ",".