Are there discrete analytic functions which are not natural functions? I think not, at least under rather lax conditions.
Consider an arbitrary expansion $f(x)=\sum_0^{\infty}a_k \binom{x}{k}$ with the $a_k$ real. It is defined for all non-negative integer $x=n$ since only the first $n+1$ terms are non-zero. As long as $\lim_{k \to \infty}\frac{a_k}{k!}=0$ we also have $f(x)$ convergent for all real $x$ and, by your definition, $f(x)$ is discrete-analytic. Then $f$ is determined by it's values at the integers.
Also, given an arbitrary sequence $y_0,y_1,y_2,\cdots$ we can uniquly make an expansion $f(x)=\sum_0^{\infty}a_k \binom{x}{k}$ with $f(m)=y_m$ by making $a_m$ whatever makes $f(m)=y_m$ given the already determined values $a_0,a_1,\cdots,a_{m-1}$ (Which amounts to the formula you gave, i.e. by repeatedly applying $\Delta$.) If we shift the values to have instead $y_n,y_{n+1},y_{n+2},\cdots$ that is just another sequence and the same procedure will work. Am I missing something?