For $\Re(s) > 1$ and by analytic continuation for $\Re(s) > -K$ $$\zeta(s) \Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^x-1}dx= \sum_{k=0}^K\frac{B_k}{k!} \frac{1}{s+k-1}+\int_0^\infty x^{s-2}(\frac{x}{e^x-1}-\sum_{k=0}^K\frac{B_k}{k!} x^k 1_{x < 1}) dx$$ ie. the Taylor expansion of $\frac{x}{e^x-1}$ gives the residues of the simple poles of $\zeta(s)\Gamma(s)$ at negative integers. They are rational numbers because the Taylor series for $\frac{e^x-1}{x}$ has rational coefficients.
For other Dirichlet series we have
$$\sum_{n=1}^\infty a_n n^{-s} \Gamma(s) = \int_0^\infty x^{s-1}\sum_{n=1}^\infty a_n e^{-nx}dx$$ but in general $x^m\sum_{n=1}^\infty a_n e^{-nx}$ has no reason to be smooth at $x=0$ nor to have rational derivatives.
I think that you misread the equation. In the paper, the rhs is $\frac{\log (x)}{m}$ while the expansion at the top of page $85$ is in terms of $\frac m{\log (x)}$.
Anyway, considering
$$-\frac{\zeta '(x)}{\zeta (x)-1}=k$$ around $x=1$, we have
$$\zeta (x)=\frac{1}{x-1}+\gamma -\gamma _1 (x-1)+\frac{1}{2} \gamma _2 (x-1)^2+O\left((x-1)^3\right)$$ So, we know the expansion of $\zeta '(x)$ and lon division gives
$$-\frac{\zeta '(x)}{\zeta (x)-1}=\frac{1}{x-1}+(1-\gamma )+$$ $$\left(2 \gamma _1+1-2 \gamma +\gamma ^2\right)
(x-1)+$$ $$\left(2 \gamma _1+(1-\gamma ) \gamma _1-2 \gamma \gamma _1-\frac{3 \gamma
_2}{2}+1-3 \gamma +3 \gamma ^2-\gamma ^3\right) (x-1)^2+O\left((x-1)^3\right)$$ which is quite good up to $x=5$.
Using series reversion (as @Gary commented)
$$x=1+\frac{1}{k}+\frac{1-\gamma }{k^2}+\frac{2 \left(\gamma _1+1-2 \gamma +\gamma
^2\right)}{k^3}+O\left(\frac{1}{k^4}\right)$$
Edit
If I had to solve the equation for $x$, I would prefer to solve instead
$$-\frac{\zeta (x)-1}{\zeta '(x)}=a \qquad \text{with} \qquad a=\frac 1k$$ which is much better conditioned.
A nonlinear regression gives (the coefficients were made rational), with $R^2 > 0.999999$, as an estimate,
$$x_0=\frac{1-\frac{65 }{92}a-\frac{82 }{273}a^2+\frac{27 }{125}a^3} { 1-\frac{171 }{101}a+\frac{88 }{95}a^2-\frac{16 }{99}a^3}$$
Some results
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
1 & 2.91932 & 2.91746 \\
2 & 1.64084 & 1.64232 \\
3 & 1.38780 & 1.38992 \\
4 & 1.27823 & 1.28026 \\
5 & 1.21697 & 1.21881 \\
6 & 1.17784 & 1.17949 \\
7 & 1.15067 & 1.15216 \\
8 & 1.13070 & 1.13205 \\
9 & 1.11541 & 1.11664 \\
10 & 1.10332 & 1.10445
\end{array}
\right)$$
Update
Another possible solution is to build the $[3,3]$ Padé approximant around $x=1$ and, making all coefficients rational have
$$f(x)=-\frac{\zeta (x)-1}{\zeta '(x)}\sim \frac {t+\frac{157 }{1308}t^2+\frac{2 }{309}t^3} {1+\frac{260 }{479}t+\frac{29 }{321}t^2+\frac{1}{224}t^3}=g(x)\qquad \text{where} \qquad t=x-1$$ which is quite good for $1 \leq x \leq 20$. To give an idea
$$\Phi=\int_0^{10} \Big[f(x)-g(x)\Big]^2\,dx=6.35 \times 10^{-6}$$
$$\Phi=\int_0^{20} \Big[f(x)-g(x)\Big]^2\,dx=4.56 \times 10^{-4}$$
Using $g(x)$, we then just need to solve the cubic equation in $t$
$$-1+\left(k-\frac{260}{479}\right) t+\left(\frac{157 k}{1308}-\frac{29}{321}\right)
t^2+\left(\frac{2 k}{309}-\frac{1}{224}\right) t^3=0$$ The discriminant is always negative and using the hyperbolic method, we have the explicit solution for $t(k)$.
Repeating the same calculations as above
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{solution} \\
1 & 2.91750 & 2.91746 \\
2 & 1.64232 & 1.64232 \\
3 & 1.38992 & 1.38992 \\
4 & 1.28026 & 1.28026 \\
5 & 1.21881 & 1.21881 \\
6 & 1.17949 & 1.17949 \\
7 & 1.15216 & 1.15216 \\
8 & 1.13205 & 1.13205 \\
9 & 1.11664 & 1.11664 \\
10 & 1.10445 & 1.10445
\end{array}
\right)$$
Best Answer
The zeta function is defined as a sum over the positive integers, but as far as actually evaluating it, it turns out to be more natural to think of it as a sum over all nonzero integers; thus we should really be thinking about $\sum_{n \neq 0} \frac{1}{n^k}$. For $k$ even this is just $2 \zeta(k)$ and there are various ways to evaluate this more symmetric sum, e.g. by writing down a meromorphic function with the right poles, or a Fourier series with the right coefficients, etc. But for $k$ odd this is equal to zero, since terms cancel with their negatives! Written in this way, the zeta function at even integers reveals its alter ego as an Eisenstein series in one dimension.
This cancellation phenomenon occurs in Euler's classic "proof," since the infinite product for $\sin z$ that he uses has zeroes at all integer multiples of $\pi$, not just the positive ones. It also occurs in the general proof that proceeds by considering the generating function $\frac{z}{e^z - 1}$ for the Bernoulli numbers. As you might know, the closed form of $\zeta(2k)$ involves Bernoulli numbers, and again $\frac{z}{e^z - 1}$ has poles at $2 \pi i n$ for all nonzero integers $n$, not just the positive ones. I describe how this works in slightly more detail here.
Another way to think about the difference between the even and odd cases is that one can think of the even cases as $L^2$ norms of appropriate Fourier series; this is precisely how a standard proof of the evaluation of $\zeta(2)$ works. But for the odd cases we don't get an $L^2$ norm; instead we get a mysterious inner product.