\begin{equation} \sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m \end{equation}
Note: $S^{(3)}_m$ belongs to the Stirling number of the first kind https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind If you see the wiki page "Table of values" $S^{(3)}_m$ corresponds to the third column.
Please let me know if addition information is needed.
The double series can be factored \begin{eqnarray*} & \sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m \\ &= \left( \sum_{m=1}^\infty \frac{S_m^{(3)}}{m! }(-1 + u)^{(m - 1)} (\frac{x}{-1 + x})^m \right) \left( \sum_{n=1}^\infty (-1)^{n } \frac{1}{ n} (-1 + u)^{ n } \right) \end{eqnarray*} The second sum is the power series for $ \ln $ and the first sum can be calculated using \begin{eqnarray*} \sum_{n=k}^{\infty} (-1)^{n-k} { n \brack k } \frac{z^n}{n!} = \frac{( \ln(1+z))^k}{k!}. \end{eqnarray*}