The RMF is definitely related to umbral calculus via the modified Mellin transform (MMT) pair and symbolic extension of the iconic Euler gamma function integral. The proof you copied? I don't know. The MMT pair allows for interpolation of the coefficients of generating functions, often directly connected to sinc and/or Newton interpolation.

First consider the MMT and its inverse

$$\tilde{f}(s) = MMT[f(x)] = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx$$

$$f(x) = MMT^{-1}[\tilde{f}(s)] = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \tilde{f}(s) \frac{x^{-s}}{(-s)!} \; ds .$$

Then the RMF holds for a class of functions such that

$$f(x) = e^{-a.x} = \sum_{n \geq 0} \frac{(-a.x)^n}{n!} = \sum_{n=0} a_n \frac{(-x)^n}{n!} =
\sum_{n=0} \tilde{f}(-n) \frac{(-x)^n}{n!} \; ,$$

that is, such that we may close the complex contour to the left (e.g., in the sense of the limit of a semicircle with its radius expanding to infinity) for $0 < \sigma < 1$ and $0 < x < 1$ when $F(s)$ has no singularities/poles within the contour. This rep allows an extension of the RMT (and the Mellin transform) to cases in which poles are present in $F(s)$ and other ranges of $x$.

Also note (see, e.g., Gelfand and Shilov's "Generalized Functions") the relation

$$D_x^{m+n+1} \; H(x) \frac{x^m}{m!} = H(x) \frac{x^{-n-1}}{(-n-1)!} = \delta^{(n)}(x)$$

reflected in the two (of several) reps of the fractional differintegro op equivalent under analytic continuation

$$\frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = \frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz = \frac{1}{2\pi i} \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz ,$$

with $H(x)$ the Heaviside step function.

So, under the conditions above,

$$\tilde{f}(-n) = \int_{0}^{\infty} f(x) \; \frac{x^{-n-1}}{(-n-1)!} \; dx = \int_{0}^{\infty} e^{-a. x} \; \delta^{(n)}(x) \; dx = a_n,$$

and this suggests the analytic continuation and relation to umbral calculus

$$\tilde{f}(s) = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx = \int_{0}^{\infty} e^{-a.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (a.)^{-s} = a_{-s}.$$

The iconic guiding example is the Euler gamma function integral rep with $(a.)^n = a_n = c^n$

$$ (a.)^{-s} = a_{-s} = c^{-s} = F(s) = MT[f(x)= e^{-c\; x}] = \int_{0}^{\infty} e^{-c \; x} \; \frac{x^{s-1}}{(s-1)!} \; dx = \frac{1}{c^{s}}.$$

Another useful example, which vividly illustrates the relation to the Appell Sheffer sequences of umbral calculus (of which the $x^n$ with e.g.f. $e^{x}$ is the basic example), is the integral rep for (what I call) the Bernoulli function, simply related to the Hurwitz zeta function and generalizing the Bernoulli polynomials,

$$ B_{-s}(z) = (B.(z))^{-s} = \int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$

$$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s+1,z)$$

where the e.g.f. for the Bernoulli polynomials with $(b.)^n = b_n$ the Bernoulli numbers is

$$e^{B.(x)t} = e^{(b.+x)t} = e^{b.t} e^{xt} = \frac{t}{e^t-1} \; e^{xt}.$$

Note that

$$B_n(z) = -n \; \zeta(1-n,z),$$

$$B_n(1) = -n \; \zeta(1-n,1) =-n \; \zeta(1-n) (Riemann) = (-1)^n B_n(0) = (-1)^n b_n.$$

Through this characterization, it is not too difficult to show that the Bernoulli function inherits all the elegant properties of a regular Appell sequence, such as $D_z \; B_{s}(z) = s \; B_{s-1}(z)$.

Riemann knew all this stuff. Ramanujan intuited it. Hardy formalized it. I stumbled across it on a journey starting from the ladder ops of QM and a brief comment by my old math prof Stallybrass about the sequence $D^{m+n} H(x) \frac{x^m}{m!}$ in his integral transforms class an eon ago.

For application to defining fractional powers of operators, see my answer and comments therein to the MO-Q "What does the inverse Mellin transform really mean?" and several of my blog posts, such as "The Creation / Raising Operators for Appell Sequences."

Added 2/5/21:

Other examples of interpolation of $a_n$ for the exponential generating funcrtion $g(t) = e^{a.t}$ from the MMT of $f(t) = g(-t) = e^{-a.t}$, or, conversely, surmising the MMT of $f(t)$ from the Taylor series coefficients of $g(t)$ via $a_n \; |_{n \rightarrow -s} =a_{-s} =\tilde{f}(s)$:

**1)** $\;g(t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \; \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = g(-t) = \cos(t) = \sum_{n \geq 0} \cos(\pi \frac{n}{2}) \; \frac{t^n}{n!} ),$

$\; \; \; \; \;\tilde{f}(s) =\cos(\pi \frac{s}{2})$ for $0 < Re(s) < 1,$

**2)** $\;g(t) = \sin(t)= \sum_{n \geq 0} \sin(\pi \frac{n}{2}) \; \frac{t^n}{n!},$

$\; \; \; \; \;f(t) = g(-t) = \sin(-t) = \sum_{n \geq 0} \sin(-\pi \frac{n}{2}) \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) =-\sin(\pi \frac{s}{2})$ for $-1 < Re(s) < 1,$

**3)** $\;g(t) = \frac{1}{1-t} = \sum_{n \geq 0} \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) =(-s)! $ for $0 < Re(s) < 1,$

**4)** $\;g(t) = \frac{1}{1+t} = \sum_{n \geq 0} \cos(\pi n) \; n! \; \frac{t^n}{n!} ,$

$\; \; \; \; \;f(t) = g(-t) = \frac{1}{1-t} = \sum_{n \geq 0} \; n! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s)=\cos(\pi s) (-s)!$ for $0 < Re(s) < 1,$

**5)** $\;g(t) = \ln(1-t) = \sum_{n \geq 0} \; -(n-1)! \; \frac{t^n}{n!} ,$

$\; \; \; \; \;f(t) = \ln(1+t) = -\sum_{n \geq 0} \cos(\pi n) \; (n-1)! \; \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) = -(-s-1)! $ for $-1 < Re(s) < 0,$

**6)** $\;g(t) =\sum_{n \ge 0} \frac{x^n}{n!} \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = J_0(2 \sqrt{xt}) =\sum_{n \ge 0} (-1)^n \frac{x^n}{n!} \frac{t^n}{n!},$

$\; \; \; \; \;\tilde{f}(s) = \frac{x^{-s}}{(-s)!}$ for $0 < Re(s) < \frac{3}{4},$

**7)** $\;g(t) = e^{-t^2} =\sum_{n \ge 0} \cos(\frac{\pi n}{2}) \; \frac{n!}{(\frac{n}{2})!} \; \frac{t^n}{n!}, $

$\; \; \; \; \;f(t) = g(-t) = e^{-t^2},$

$\; \; \; \; \;\tilde{f}(s) = \cos(\pi\frac{ s}{2}) \; \frac{(-s)!}{(-\frac{s}{2})!} = \frac{1}{2}\frac{(\frac{s}{2}-1)!}{(s-1)!} \;$ for $ Re(s) > 0.$

**8)** $\;g(t) =\sum_{n \ge 0} \; \frac{a_{\bar{n}}\; b_{\bar{n}}}{c_{\bar{n}}} \; \frac{t^n}{n!} = F(a,b;c;t)$, the hypergeometric function, where, e.g.,

$\; \; \; \; \;a_{\bar{n}} = \frac{(a+n-1)!}{(a-1)!}$, the rising factorial,

$\; \; \; \; \;f(t) = g(-t) = F(a,b;c;-t),$

$\; \; \; \; \;\tilde{f}(s) = \frac{a_{-\bar{s}} \; b_{-\bar{s}}}{c_{-\bar{s}}},$
see the Mellin-Barnes contour integral.

1more comment