I want to show that $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\Big( \frac{x^{a-1}}{\sinh x} - x^{a-2}\Big) \, \mathrm dx \, , \quad 0 < \Re (a) <1. \tag{1}$$
I can show that $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty} \frac{x^{a-1}}{\sinh x} \, \mathrm dx \, , \quad \Re (a) >1, \tag{2}$$
but I don't know if $(2)$ can be used to prove $(1)$.
EDIT:
Following the approach used in this paper to show that $$\Gamma(s) \zeta(s) = \int_{0}^{\infty} \left(\frac{1}{e^{x}-1}-\frac{1}{x} \right)x^{s-1} \, \mathrm dx \, , \quad0 < \Re(s) < 1, $$
subtract $$\frac{\Gamma(a)}{1-a} b^{1-a} = \int_{0}^{\infty} e^{-b x}x^{a-2} \, \mathrm dx \, , \quad \left( \Re(a) >1, \, b>0 \right),$$ from $(2)$ to get
$$\Gamma(a) \left(2 \zeta(a) \left(1-\frac{1}{2^{a}} \right) - \frac{b^{1-a}}{1-a} \right) = \int_{0}^{\infty} \left(\frac{1}{\sinh x} - \frac{e^{-bx}}{x} \right)x^{a-1} \, \mathrm dx \, , \quad \Re(a) > 0.$$
What this did is compensate for the fact that $\frac{1}{\sinh z}$ has a simple pole at $z=0$ with residue $1$.
Now restrict $a$ to the vertical strip $0 < \Re (a) <1$, and let $b$ tend to zero to get $$ 2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\left( \frac{1}{\sinh x} - \frac{1}{x}\right)x^{a-1} \, \mathrm dx \ , \quad 0 < \Re(a) <1.$$
You could then appeal to the identity theorem to argue that $(1)$ actually holds for $-1 < \Re(a) <1$.
First step, write the integral as $$ I = \int_{0}^{\infty}\left(-{\frac {{x}^{a-2}{{\rm e}^{2\,x}}}{{{\rm e}^{2\,x}}-1}}+2\,{\frac {{x }^{a-1}{{\rm e}^{x}}}{{{\rm e}^{2\,x}}-1}}+{\frac {{x}^{a-2}}{{{\rm e} ^{2\,x}}-1}}\right) dx. $$
Second step, use the change of variables $ 2x=u $. Third step, use the identity (the Hurwitz zeta function)
$$ \zeta(s,q)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.$$
Here are related problems: I, II. By the way, I tried Maple and WolframAlpha, but no answer.