I believe the $(-1)^{n-s}$ term in your binomial sum should just be $(-1)^n$ as in $$\left(\frac{t}{e^t-e^{-t}}\right)^s=\sum\limits_{n=0}^{\infty} \binom{-s}{n}\, (-1)^n\, t^s\, e^{-(2 n+s) t}$$ and since $$-\frac{2^s}{s} \int_0^{\infty } t^s\, e^{-(2 n+s) t} \, dt=-2^s\, \Gamma(s)\, (2 n+s)^{-s-1}$$ I believe the Mellin transform result should actually be $$-2^s\, \Gamma(s)\, \sum\limits_{n=0}^{\infty} \binom{-s}{n}\, (-1)^n\, (2 n+s)^{-s-1}$$ and consequently it would appear your subsequent inverse Mellin transform is wrong.

@StevenClark If you click on the Mathematica code and use $\frac{\Gamma(s+1)}s=\Gamma(s)$ you get the same result as after “should actually be” in your comment. Right? There are likely typos when expanding via binomial theorem; the $(-1)^{n-s}$ was supposed to be from raising $(-e^{-t})$ to the $-s-n$ power. Do you have an idea on how to find a closed form for $g(s)$, using the RM theorem etc?

I don't understand the early part of your derivation. You define $f(x)$ as the inverse of $x\, \text{csch}(x)$ and then you indicate the substitution $t\to f(t)$, but you actually use the substitution $t\to t\, \text{csch}(t)$. It appears you have the limits reversed after this substitution as $t=0$ and $t=1$ before the substitution corresponds to $t=\infty$ and $t=0$ respectively after the substitution.

Does it have to be done by using the Ramanujan master theorem? Or you are only interested in a closed form for the $\operatorname{sinhc}^{-1}$

@StevenClark Thanks. It likely should have been written as $f(t)\to t$ in the question

@Marco There is likely no closed form for the inverse, but if you have it, then you can post it. The reasoning for the Ramanujan master theorem is that it also uses a Mellin transform like in the question, but instead of the question’s “integral of a series” expression of the inverse, you would get a “series of a series” solution for the inverse with the theorem, which looks cleaner. A closed form for the Mellin transform would also give a cleaner expression than in the question

@TymaGaidash I was thinking about Lagrange inversion formula, but $\operatorname{sinhc}$ isn't actually analytic near the origin, so i don't think it works

@Marco Actually you can expand it as $x+\sum\limits_{n=1}^\infty\frac1{n!}\frac{d^{n-1}}{dx^{n-1}}\left(x-\frac{\sinh(x)}x\right)^n$. Expanding the derivatives could give a double/triple sum which is longer than the expression in the question and has a small radius of convergence.

I don't understand your notion of applying Ramanujan's master theorem which is used to derive the Mellin transform from a series representation of a function. Is your notion to derive $\mathcal{M}_x[f(x)](s)=\int\limits_0^{\infty} f(s)\, x^{s-1} \, dx$ by applying Ramanujan's master theorem to a series representation of $f(t)$ derived via series reversion (Mathematica InverseSeries function) of a series representation for $t\, csch(t)$?

@StevenClark You may notice that from the statement of the theorem the series coefficients are $\varphi(s)=\frac1{\Gamma(-s)}\int_0^\infty x^{-s-1} f(x)dx$, so that $f(x)=\sum\limits_{n=0}^\infty \frac{\varphi(n)}{n!}(-x)^n,\varphi(n)=\frac1{\Gamma(-n)}\mathcal M_x[f(x)](-n)$. However, trying the Mellin transform in the question makes $\varphi(n)=0$ due to the $\Gamma(-n)$. Therefore, someone else who knew more about it could have more success than here.

It seems to work for $$\mathcal{M}_x[x\, \text{csch}(x)](s)=\left(2-2^{-s}\right) \Gamma(s+1)\, \zeta(s+1)=\Gamma(s)\, \phi(-s)$$ because in $$\phi(s)=\frac{\left(2-2^s\right) \Gamma(1-s)\, \zeta(1-s)}{\Gamma(-s)}$$ the poles of $\Gamma(1-s)$ in the numerator cancel the poles of $\Gamma(-s)$ in the denominator, but you have to evaluate it as $$\phi(n)=\underset{s\to n}{\text{lim}}\frac{\left(2-2^s\right) \Gamma(1-s)\, \zeta (1-s)}{\Gamma (-s)}.$$

After thinking about it some more, I don't believe Ramanujan's master theorem will be useful here at its based on a power series around $x_o=0$ and $f(x)$ doesn't converge at $x=0$ which seems to imply such a power series doesn't exist.

@StevenClark That is understandable. Do you have another idea?

For the Mellin transform $$F(s)=\mathcal{M}_{\tau}[f(\tau)](s)=\int\limits_0^1 f(\tau)\, \tau^{s-1} \, d\tau,$$ I believe the substitution $\tau\to g(t)=t\, \text{csch}(t)$ leads to $$F(s)=\int_{\infty}^0 f(g(t))\, g(t)^{s-1} \, dg(t)=\int_{\infty}^0 t\, g(t)^{s-1}\, g'(t) \, dt.$$ I mentioned you had the limits reversed in an earlier comment as $\tau=0$ and $\tau=1$ before the substitution correspond to $t=\infty$ and $t=0$ respectively after the substitution.

So I believe you have sign errors in your first two formulas for $M_t(f(t))$ which you seem to have compensated for in the linked Mathematica code. Your formula for $\operatorname{sinhc}^{-1}(x)$ initially indicates $\pm$, but then once again corrects for the sign error. While its true $t\, \text{csch}(t)$ is an even function of $t$, the inverse Mellin transform of $\mathcal{M}_x[f(\tau)](s)$ should return the positive solution, not the negative solution.

I've been thinking about deriving a power series for $f(t)$ via reversion of a power series for $t\, \text{csch}(t)$, but since $t\, \text{csch}(t)$ has poles at $2 \pi i c\,,\ c\in\mathbb{Z}\land c\ne 0$, there is no power series for $t\, \text{csch}(t)$ that converges for $0<t<\infty$. Therefore a power series for $f(t)$ derived via reversion of a power series for $t\, \text{csch}(t)$ would not converge for $0<t<1$. For example, the MacClaurin series $$t\, \text{csch}(t)=1-\sum\limits_{n=1}^{\infty} \frac{\left(2^{2 n}-2\right) B_{2 n}}{(2 n)!} t^{2 n}$$ has a radius of convergence of $\pi$.

I posted the results of my investigation below which perhaps provides some insight with respect to a closed-form representation for $F(s)=\mathcal{M}_t[f(t)](s)$.

@StevenClark While there may be a series reversion, which can be posted in this question, the goal of the question is to simplify/re-express the boxed formula for which reasons are given in the question.

I was thinking of a power series for $f(t)$ in the context of deriving an alternative representation of $F(s)=\mathcal{M}_t[f(t)](s)$ to compare against your derived formula for $F(s)=\mathcal{M}_t[f(t)](s)$ to serve as a double-check on the correctness of your derivation. But I believe the answer to the question you linked is no for the reason I pointed out.

This question of mine contains a discussion of the inverse of sinh(x)/x by Claude Leibovici: math.stackexchange.com/questions/3557767/…