Here is some observations: Let
$$U(x) = \sum_{n=1}^{\infty} u_n x^n. $$
From the identity $\displaystyle -\frac{\log(1-x)}{x} = \sum_{n=0}^{\infty} \frac{x^n}{n+1}$, it follows that
$$ -\frac{\log(1-x)}{x} U(x) = \sum_{n=1}^{\infty} \left( \sum_{k=1}^{n} \frac{u_k}{n+1-k} \right) x^n = \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$
and hence
$$ U(x) = -\frac{x^2}{(1-x)\log(1-x)}. $$
Now let
$$ -\frac{x}{\log(1-x)} = 1 - \sum_{n=1}^{\infty} a_n x^n. $$
Then it is plain to confirm that
$$ u_n = 1 - \sum_{k=1}^{n-1} a_k. $$
Regarding the behavior of $(a_n)$, we claim the following assertions:
- $a_n > 0$ for all $n = 1, 2, 3, \cdots$ and
- $\sum a_n = 1$.
Once $a_n > 0$ is established, then the second assertion immediately follows. Indeed, the sum $\sum a_n$ tends to some limit in $[0, \infty]$. Then a simple version of Abelian theorems shows that
$$ \sum_{n=1}^{\infty} a_n = \lim_{x\to 1^-} \sum_{n=1}^{\infty} a_n x^n = \lim_{x\to 1^-} \left( 1 + \frac{x}{\log (1-x)} \right) = 1.$$
Thus it remains to prove the first assertion. Let
$$ f(x) = \frac{x-1}{\log x}. $$
Then it is easy to check that $f(x) \geq 0$ on $[0, \infty)$ and
$$ a_n = \frac{(-1)^{n-1}}{n!} f^{(n)}(1). $$
This shows that it suffices to check the condition $ (-1)^{n-1} f^{(n)}(x) > 0 $ for $x > 0$ and $n = 1, 2, 3, \cdots$. In other words, it is sufficient to show that $f$ is a Bernstein function. This follows once we show that
$$ f(s) = \int_{0}^{\infty} \left( -\int_{0}^{1}\frac{t^{u-2}}{(u-2)!} \, du \right) \left( 1 - e^{-st} \right) \, dt, \tag{1}$$
since the integrand is positive inside the domain of the integration. But this follows from the following calculation:
\begin{align*}
\frac{s-1}{\log s}
&= s \int_{0}^{1} \frac{du}{s^u}
= s \int_{0}^{1} \frac{1}{\Gamma(u)} \int_{0}^{\infty} t^{u-1} e^{-st} \, dtdu \\
&= s \int_{0}^{\infty} \left( \int_{0}^{1} \frac{t^{u-1}}{\Gamma(u)} \, du \right) e^{-st} \, dt \\
&= \left[ \left( \int_{0}^{1} \frac{t^{u-1}}{\Gamma(u)} \, du \right) \left( 1 - e^{-st} \right) \right]_{0}^{\infty} - \int_{0}^{\infty} \left( \int_{0}^{1} \frac{t^{u-1}}{\Gamma(u)} \, du \right)' \left( 1 - e^{-st} \right) \, dt.
\end{align*}
Now combining the two assertions and $(1)$, we have
\begin{align*}
u_n
= \sum_{k=n}^{\infty} a_k
&= \sum_{k=n}^{\infty} \frac{(-1)^{k-1}}{k!} f^{(k)}(1) \\
&= \sum_{k=n}^{\infty} \int_{0}^{\infty} \int_{0}^{1} \frac{1-u}{\Gamma(u)} \frac{t^{u+k-2}}{k!} e^{-t} \, du \, dt \\
&= \int_{0}^{\infty} \int_{0}^{1} \frac{1-u}{\Gamma(u)} t^{u-2} \int_{0}^{t} \frac{v^{n-1}}{(n-1)!} e^{-v} \, dv \, du \, dt \\
&= \int_{0}^{\infty} \int_{0}^{1} \frac{t^{u+n-2}}{(n-1)!\Gamma(u)}e^{-t} \, du \, dt \\
&= \int_{0}^{1} \frac{\Gamma(u+n-1)}{(n-1)!\Gamma(u)} \, du \\
&= \int_{0}^{1} \prod_{k=1}^{n-1} \left( 1 - \frac{u}{k} \right) \, du.
\end{align*}
Finally, heuristic calculation then suggests that
$$\lim_{n\to\infty} u_n \log n = 1, $$
which I'm trying to prove.