This is not a full answer as it does not really provide a "closed form" representation of the series, though I highly suspect no such closed form exists. Rather, here is short discussion on some different forms of this and a related series that could be of some interest to the asker.
Firstly, we have the following easy to prove identity
\begin{equation}
\boxed{\sum_{k=2}^\infty \zeta'(k)x^k=\sum_{n=2}^\infty\frac{\log(n)x^2}{n(n-x)}}
\end{equation}
for $|x|<2$ which provides the analytic continuation of the first sum to the whole complex plane except for integers $x\geq 2$.
To see this, write
\begin{equation}
\begin{split}
\sum_{k=2}^\infty \zeta'(k)x^k&=\sum_{k=2}^\infty\left[\sum_{n=2}^\infty\frac{\log(n)}{n^k}\right]x^k\\
&=\sum_{n=2}^\infty\log(n)\sum_{k=2}^\infty\left(\frac{x}{n}\right)^k\\
&=\sum_{n=2}^\infty\frac{\log(n)x^2}{n(n-x)}\\
\end{split}
\end{equation}
The related series representing the even part of the series discussed above admits the following representation
\begin{equation}
\boxed{\sum_{n=1}^\infty\zeta'(2n)y^{2n}\!=\!\frac1{2}[\gamma+\log(2\pi)](1-\pi y\cot(\pi y))+\pi^2y^2\csc(\pi y)\!\!\int_0^1\!\!\!\sin(\pi y(2x-1))\log\Gamma(x)dx}
\end{equation}
for $|y|<2$ (for $y\in\{-1,0,1\}$, the right-hand expression can be interpreted as a limit). Note that the integral converges for any complex $y$, so this identity again provides an analytic continuation of the sum to the whole complex plane except for integers $y$ with $|y|\geq 2$.
We begin with the following well known Fourier series
\begin{equation}
\begin{split}
\log\Gamma(x)&=\frac12\log(2\pi)+\sum_{k=1}^\infty \frac{1}{2k}\cos(2k\pi x)+\sum_{k=1}^\infty \frac{1}{\pi k}[\gamma+\log(2k\pi)]\sin(2k\pi x)\\
B_{2n-1}(x)&=2(2n-1)!(-1)^{2n-1}\sum_{k=1}^\infty \frac{\sin(2k\pi x)}{(2k\pi)^{2n-1}}\\
\end{split}
\end{equation}
for $0<x<1$ and $n\geq 1$, where $B_m(x)$ is the $m$-th Bernoulli polynomial. Now, by using the above series and appealing to the orthogonality relationships between sin and cos, we may show that.
\begin{equation}
\int_0^1 B_{2n-1}(x)\log\Gamma(x)dx=2(-1)^n\frac{(2n-1)!}{(2\pi)^{2n}}[(\gamma+\log(2\pi))\zeta(2n)-\zeta'(2n)]
\end{equation}
for $n\geq 1$. Finally, using the generating function identity
$$\frac{t \sinh((x-1/2)t)}{2\sinh(t/2)}=\sum_{n=1}^\infty B_{2n-1}(x)\frac{t^{2n-1}}{(2n-1)!}$$
we have that
\begin{equation}
\begin{split}
\int_0^1 \frac{t\sinh((x-1/2)t)}{2\sinh(t/2)}\log\Gamma(x)dx&=2\sum_{n=1}^\infty (-1)^n[(\gamma+\log(2\pi))\zeta(2n)-\zeta'(2n)]\frac{t^{2n-1}}{(2\pi)^{2n}}\\
&=[\gamma+\log(2\pi)]\left(\frac1{t}+\frac{i}{2}\cot\left(\frac{t}{2i}\right)\right)-\frac2{t}\sum_{n=1}^\infty\zeta'(2n)\left(\frac{t}{2\pi i}\right)^{2n}\\
\end{split}
\end{equation}
Performing the change of variable $t\mapsto 2\pi iy$, we have that
\begin{equation}
\int_0^1 \frac{\pi i y \sin(\pi y(2x-1))}{\sin(\pi y)}\log\Gamma(x)dx=[\gamma+\log(2\pi)]\left(-\frac{i}{2\pi y}+\frac{i}{2}\cot(\pi y)\right)+\frac{i}{\pi y}\sum_{n=1}^\infty\zeta'(2n)y^{2n}
\end{equation}
taking the imaginary part of both sides, and rearranging, we have that
\begin{equation}
\sum_{n=1}^\infty\zeta'(2n)y^{2n}\!=\!\frac1{2}[\gamma+\log(2\pi)](1-\pi y\cot(\pi y))+\pi^2y^2\csc(\pi y)\!\!\int_0^1\!\!\!\sin(\pi y(2x-1))\log\Gamma(x)dx
\end{equation}
as desired.
