In this question I overlooked that the class $f_s(x)=e^{sT(x)}$ for $T(x)=\frac{1}{\log x},$ is Schwartz, on $x \in (0,1).$ Therefore we can obtain a functional equation, see this answer.
However, I am having some difficulties obtaining the Gamma factor. I started by writing down the calculation:
$$\Gamma(f,r,s)=\int_{\mathbb R^\times \cap ~(0,1)} |x|^r~f_s(x)~{dx\over |x|}=2 \sqrt{\frac{r}{s}}K_1(2\sqrt{r s})$$
Where $K_1$ is a modified Bessel. But something isn't right. It would seem that I have to find the Fourier transform of $f_s(x).$ However, I'm not sure how to get a closed form for that.
How do you get the correct Gamma factor for the class of Schwartz functions $f_s(x)?$
This Schwartz function is its own inverse i.e. satisfies $f\circ f=\mathrm{id}$ but I'm not sure if that can help here.
Edit: Someone pointed me in the direction of Pontryagin duality. They said to view $(0,1)$ as the torus $\mathbb{R}/\mathbb{Z}$ with group operation as addition modulo $1$. And they said that roughly speaking, the Fourier transform of a function on $(0,1)$ is a Fourier series.
I am currently following this lead and hopefully it will resolve this question.
Consider the even function
$$f_r(x)=e^{\frac{r}{\log(|x|)}},\quad-1<x<1\tag{1}$$
where
$$\Gamma\left[f_r,s\right]=\int\limits_{-1}^1 f_r(x)\, |x|^{s-1}\, dx=2 \int\limits_0^1 f_r(x)\, x^{s-1}\, dx=4\, \sqrt{\frac{r}{s}}\, K_1\left(2\, \sqrt{r s}\right),\quad r>0\land\Re(s)>0\tag{2}.$$
Most of the remainder of this answer will focus on the case
$$f_1(x)=e^{\frac{1}{\log(|x|)}},\quad-1<x<1\tag{3}$$
Since $f_1(x)$ is an even function of $x$ it can be approximated on the interval $x\in(-1,1)$ by the 2-periodic Fourier cosine series
$$f_1(x)\approx \frac{a_0}{2}+\sum\limits_{k=1}^K a_k\, \cos(\pi k x)\tag{4}$$
where
$$a_k=\int\limits_{-1}^1 f_1(x)\, \cos(\pi k x)\, dx\tag{5}$$
I don't have a closed form for the Fourier series coefficient $a_k$ defined in formula (4) above, but these coefficients can be approximated using numeric integration and decrease in magnitude as the value of $k$ increases as illustrated in the following table.
$$\begin{array}{cc} k & a_k \\ 0 & 0.559464 \\ 1 & 0.342683 \\ 2 & 0.0618051 \\ 3 & 0.0188978 \\ 4 & 0.0256676 \\ 5 & 0.0115177 \\ 6 & 0.0113023 \\ 7 & 0.00845802 \\ 8 & 0.00675756 \\ 9 & 0.00606805 \\ 10 & 0.00487649 \\ 11 & 0.00449404 \\ 12 & 0.00381808 \\ 13 & 0.00349118 \\ 14 & 0.00310478 \\ 15 & 0.00282947 \\ 16 & 0.00258617 \\ 17 & 0.00236765 \\ 18 & 0.00219612 \\ 19 & 0.0020276 \\ 20 & 0.00189614 \\ \end{array}$$
Figure (1) below illustrates the Fourier series for $f_1(x)$ defined in formula (4) above in orange overlaid on $f_1(x)$ defined in formula (3) above in blue where the Fourier series is evaluated at $K=20$.
Figure (1): Illustration of Fourier series for $f_1(x)$ (orange) overlaid on $f_1(x)$ (blue)
The Fourier transform
$$\hat{f_1}(w)=\int\limits_{-1}^1 f_1(x)\, e^{-i 2 \pi w x}\tag{6}$$
of $f_1(x)$ can be approximated by the following term-wise Fourier transform
$$\hat{f_1}(w)\approx \frac{a_0}{2}\frac{\sin(2 \pi w)}{\pi w}+\sum\limits_{k=1}^K a_k\, \frac{2 k \sin(\pi k) \cos(2 \pi w)-4 w \cos(\pi k) \sin(2 \pi w)}{\pi k^2-4 \pi w^2}\tag{7}$$
of formula (4) above.
Figure (2) below illustrates the formula (7) series above for $\hat{f_1}(w)$ in orange overlaid on formula (6) for $\hat{f_1}(w)$ in blue where the formula (7) series is evaluated at $K=20$ and numeric integration is used to evaluate the Fourier transform in formula (6) above.
Figure (2): Illustration of formula (7) series for $\hat{f_1}(w)$ (orange) overlaid on $\hat{f_1}(w)$ (blue)
The function
$$\Gamma\left[\hat{f_1},1-s\right]=\int\limits_{-\infty}^\infty \hat{f_1}(w)\, |w|^{-s}\, dw=2 \int\limits_0^\infty \hat{f_1}(w)\, w^{-s}\, dw\tag{8}$$
can be approximated by the following term-wise Mellin transform
$$\Gamma\left[\hat{f_1},1-s\right]\approx \frac{a_0}{2} \left(-2^{s+1} \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(-s)\right)+\frac{1}{2} \pi^{s+\frac{1}{2}} \sec\left(\frac{\pi s}{2}\right) \sum\limits_{k=1}^K a_k\, \left(\pi k \sin (\pi k) \, _1\tilde{F}_2\left(1;\frac{s+2}{2},\frac{s+3}{2};-\frac{1}{4} k^2 \pi ^2\right)+2 \cos (\pi k) \, _1\tilde{F}_2\left(1;\frac{s+1}{2},\frac{s+2}{2};-\frac{1}{4} k^2 \pi ^2\right)\right)\tag{9}$$
of formula (7) above.
Figure (3) below illustrates $\Gamma\left[f_1,s\right]\, \zeta(s)$ in blue and $\Gamma\left[\hat{f_1},1-s\right]\, \zeta(1-s)$ in orange where formula (2) above is used to evalute $\Gamma\left[f_1,s\right]$ and formula (9) above evaluated at $K=20$ is used to approximate $\Gamma\left[\hat{f_1},1-s\right]$. Note $\Gamma\left[\hat{f_1},1-s\right]\, \zeta(1-s)$ (orange) seems to approximate $\Gamma\left[f_1,s\right]\, \zeta(s)$ (blue) for $s>0$ where the imaginary part of $\Gamma\left[f_1,s\right]$ has a branch.
Figure (3): Illustration of $\Gamma\left[f_1,s\right]\, \zeta(s)$ in blue and $\Gamma\left[\hat{f_1},1-s\right]\, \zeta(1-s)$ in orange
Figure (4) below illustrates $2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\, \Gamma\left[f_1,s\right]$ in blue and $\Gamma\left[\hat{f_1},1-s\right]$ in orange where formula (2) above is used to evaluate $\Gamma\left[f_1,s\right]$ and formula (9) above evaluated at $K=20$ is used to approximate $\Gamma\left[\hat{f_1},1-s\right]$. Note $\Gamma\left[\hat{f_1},1-s\right]$ (orange) seems to approximate $2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\,\Gamma\left[f_1,s\right]$ (blue) for $s>0$ where the imaginary part of $\Gamma\left[f_1,s\right]$ has a branch.
Figure (4): Illustration of $2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\, \Gamma\left[f_1,s\right]$ in blue and $\Gamma\left[\hat{f_1},1-s\right]$ in orange
In summary, for $f_r(x)$ defined in formula (1) above, I believe the closed form for $\Gamma\left[\hat{f_r},1-s\right]$ is
$$\Gamma\left[\hat{f_r},1-s\right]=\frac{\zeta(s)}{\zeta(1-s)}\, \Gamma\left[f_r,s\right]=2^s\, \pi^{s-1}\, \sin\left(\frac{\pi s}{2}\right)\, \Gamma(1-s)\, \Gamma\left[f_r,s\right]\tag{10}$$
where $\Gamma\left[f_r,s\right]$ is defined in formula (2) above.