(This is a revamping of MSE-Q, which has examples of Newton series interpolation for particular functions and the relation to the Mellin transform.)

With $ \; \; \displaystyle \bigtriangledown^{s}_n \; c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s}{n} \; c_n \; \; $ and $\; \; \displaystyle m_j= \int_{0}^{\infty} h(x) \; x^j \; dx \; \; $, formally

$$g(s)=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{m_j}{j!}=\int_0^\infty h(x)\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!} dx=\int_0^\infty h(x) \frac{x^{s-1}}{(s-1)!} \; dx,$$

so since $ \; \; \displaystyle m_j=j!\; g(j+1)=\int_0^\infty h(x) x^j \; dx \; \;$,

$$ g(s+1)=\bigtriangledown^{s}_{n} \bigtriangledown^{n}_{j} g(j+1).$$

Identifying $g(x+1)$ with $f(x)$ in your formula, you can see that your Newton series is the Mellin transform in disguise.

Performing the modified inverse Mellin transform

$h(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} g(s) \frac{x^{-s}}{(-s)!} ds$ .

For the Fourier transform, the analogous formula is a sinc function interpolation of a bandlimited function

$$\hat{g}(x) = \int_{-W/2}^{W/2} h(\omega) e^{i 2 \pi x \omega} d\omega.$$

Then

$$\hat{g}(x) = \sum_{n=-\infty}^{\infty} \frac{sin[\pi W\; (n/W-x)]}{
\pi W\;(n/W-x)} \hat{g}(n/W)$$

with $$e^{i2 \pi x \omega}=\sum_{n=-\infty}^{\infty} \frac{sin[\pi W \; (n/W-x)]}{\pi W\; (n/W-x)} e^{i2 \pi n \omega /W} $$

for $-W/2<\omega<W/2$ being the analogue of

$$ \frac{x^{s-1}}{(s-1)!}=\bigtriangledown_{n}^{s-1}\bigtriangledown_{j}^{n} \frac{x^j}{j!}$$

valid for $Real(s)>0$. (This is the basis for the Whittaker-Shannon theorem.)

An aside on a related Newton series and Ramanujan's Master Formula:

Using umbral notation (MSE-Q) for the Taylor series, $f(t)=e^{a.t}$, the normalized Mellin transform formally gives

$$g(s)= \displaystyle \int_{0}^{\infty} e^{-a.t} \frac{t^{s-1}}{(s-1)!}dt$$
$$ = \int_{0}^{\infty} e^{-t}\; e^{(1-a.)t} \frac{t^{s-1}}{(s-1)!}dt =\sum_{m=0}^{\infty} (-1)^m \binom{-s}{m}\sum_{k=0}^m(-1)^k \binom mk a_k \; .$$

Then $f(t)=e^{a.t}$ with $a_n=g(-n)$ since, for $s=-n$, the binomial transform here is an involution and the regularization of the normalized Mellin transform (MT) gives the Dirac delta function or its derivatives in the integrand. Both sides of the equation would converge simultaneously only over restricted values of $s$ as noted in the MO-Q link, but could be analytically continued.

(Use $\binom{s+m-1}{m}=(-1)^m \binom{-s}{m}$ for the MT term-by-term of the Taylor series for $e^{(1-a.)t)}$.)

This is the essence of Ramanujan's Master Formula/Theorem (cf. MO-Q).

So, the Newton interpolation is more akin to the Fourier transform itself; in fact, you can regard the MT here as the interpolation of the coefficients of the Taylor series with the Newton series as one of its avatars.

The inverse MT gives, for appropriate $ \sigma$ when convergent,

$$\displaystyle \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(s) \frac{(-t)^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(-n) \frac{t^{n}}{n!} = f(t) \; .$$