The main idea is that, near the largest point of the integrand (which occurs at $x=0$), we have
$$
\frac{1}{\log x} + \frac{1}{1-x} \approx \frac{1}{\log x} + 1.
$$
So we split the integral up as
$$
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx = \left[\int_0^{1/e} + \int_{1/e}^1 \right] \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx.
$$
For the second integral we have the bound
$$
0 < \int_{1/e}^{1} \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx < (e-1)^{-n}
$$
and for the first integral we write
$$
\begin{align}
\int_0^{1/e} \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \int_0^{1/e} \exp\left\{n \log\left(\frac{1}{\log x} + \frac{1}{1-x}\right)\right\}dx \\
&= \int_0^{1/e} \exp\left\{n \log\left(\frac{1}{\log x} + 1\right) + n f(x)\right\}dx,
\end{align}
$$
where
$$
f(x) = \log\left(1+\frac{x}{\left(\frac{1}{\log x}+1\right)(1-x)}\right) = O(x) \quad \text{as $x \to 0$.}
$$
We therefore expect that the largest contribution comes from a neighborhood of radius $1/n$ of the point $x=0$. The rest of the integral is bounded by
$$
0 < \int_{1/n}^{1/e} \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx < \left(-\frac{1}{\log n} + \frac{n}{n-1}\right)^n = O(e^{-n/\log n}).
$$
For $0 \leq x \leq 1/n$ we have
$$
\exp\{nf(x)\} = 1 + O(nx),
$$
so, after combining our estimates, we arrive at
$$
\begin{align}
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \int_0^{1/n} \exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx \\
&\qquad + O\left(n \int_0^{1/n} x \exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx\right)\\
&\qquad + O(e^{-n/\log n}).
\end{align}
$$
Now we make the change of variables $-y = \log\left(\frac{1}{\log x} + 1\right)$, transforming our integrals into
$$
\begin{align}
&\int_0^{1/n} \exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx \\
&\qquad = \int_0^{-\log(1-1/\log n)} e^{-ny} \exp\left\{\frac{1}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2}\,dy
\end{align}
$$
and
$$
\begin{align}
&\int_0^{1/n} x\exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx \\
&\qquad = \int_0^{-\log(1-1/\log n)} e^{-ny} \exp\left\{\frac{2}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2}\,dy.
\end{align}
$$
We can expand the factor which is independent of $n$ as
$$
\begin{align}
\exp\left\{\frac{1}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2} &= \exp\left\{-\frac{1}{y}-\frac{1}{2}\right\} \sum_{j=-2}^{\infty} c_j y^j \\
&= \exp\left\{-\frac{1}{y}-\frac{1}{2}\right\} \left( \frac{1}{y^2} -\frac{1}{12 y} - \frac{23}{288} + \frac{427 y}{51840} + \cdots \right),
\end{align}
$$
so on truncating the expansion at $j=J-1$ we obtain
$$
\begin{align}
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \sum_{j=-2}^{J-1} \frac{c_j}{\sqrt{e}} \int_0^{-\log(1-1/\log n)} e^{-ny} e^{-1/y} y^j \,dy \\
&\qquad + O\left(\int_0^{-\log(1-1/\log n)} e^{-ny} e^{-1/y} y^J \,dy\right) \\
&\qquad + O\left(n \int_0^{-\log(1-1/\log n)} e^{-ny} e^{-2/y} y^{-2} \,dy \right) \\
&\qquad + O(e^{-n/\log n}).
\end{align}
$$
All that remains is to reattach the tails of the integrals. By doing so we add an error of no more than
$$
\begin{align}
&\int_{-\log(1-1/\log n)}^\infty e^{-ny} e^{-1/y} y^j \,dy \\
&\qquad = \int_{-\log(1-1/\log n)}^\infty e^{-ny/2} \left(e^{-ny/2} e^{-1/y} y^j\right) \,dy \\
&\qquad \leq \sqrt{\int_{-\log(1-1/\log n)}^\infty e^{-ny}\,dy} \cdot \sqrt{\int_{-\log(1-1/\log n)}^\infty e^{-ny} e^{-2/y} y^{2j}\,dy} \\
&\qquad = \sqrt{\frac{2}{n}} \left(1-\frac{1}{\log n}\right)^{n/2} \sqrt{\int_{-\log(1-1/\log n)}^\infty e^{-ny} e^{-2/y} y^{2j}\,dy} \\
&\qquad < \sqrt{\frac{2}{n}} \left(1-\frac{1}{\log n}\right)^{n/2} \sqrt{\int_{0}^\infty e^{-ey} e^{-2/y} y^{2j}\,dy} \\
&\qquad = O\left(n^{-1/2} e^{-n/(2\log n)}\right)
\end{align}
$$
for $n > e$ by the Cauchy-Schwarz inequality, with the same bound when $e^{-1/y}$ is replaced with $e^{-2/y}$. Thus we have
$$
\begin{align}
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \sum_{j=-2}^{J-1} \frac{c_j}{\sqrt{e}} \int_0^\infty e^{-ny} e^{-1/y} y^j \,dy \\
&\qquad + O\left(\int_0^\infty e^{-ny} e^{-1/y} y^J \,dy\right) \\
&\qquad + O\left(n \int_0^\infty e^{-ny} e^{-2/y} y^{-2} \,dy \right) \\
&\qquad + O\left(n^{-1/2} e^{-n/(2 \log n)}\right).
\end{align}
$$
To normalize these integrals a little bit we can substitute $y = u/\sqrt{n}$ to get
$$
\begin{align}
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \sum_{j=-2}^{J-1} \frac{c_j}{\sqrt{e}} n^{-(j+1)/2} \int_0^\infty e^{-\sqrt{n}(u+1/u)} u^j \,du \\
&\qquad + O\left(n^{-(J+1)/2} \int_0^\infty e^{-\sqrt{n}(u+1/u)} u^J \,du\right) \\
&\qquad + O\left(n^{3/2} \int_0^\infty e^{-\sqrt{n}(u+2/u)} u^{-2} \,du \right) \\
&\qquad + O\left(n^{-1/2} e^{-n/(2 \log n)}\right).
\end{align}
$$
With the basic Laplace method estimate
$$
\int_0^\infty e^{-\sqrt{n}(u+\alpha/u)} u^j \,du \sim \alpha^{1/4} \sqrt{\frac{\pi}{n}} e^{-2\sqrt{\alpha n}}
$$
we may conclude that the series we've written is actually an asymptotic series, i.e.
$$
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx \sim \sum_{j=-2}^{\infty} \frac{c_j}{\sqrt{e}} n^{-(j+1)/2} \int_0^\infty e^{-\sqrt{n}(u+1/u)} u^j \,du.
$$
Here we recognize that
$$
\int_0^\infty e^{-\sqrt{n}(u+1/u)} u^j \,du = 2 K_{-j-1}\left(2\sqrt{n}\right),
$$
where $K_\nu$ is the modified Bessel function of the second kind (see eq. 10.32.10, DLMF).
One asymptotic series for our integral is thus
$$
\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx \sim \frac{2}{\sqrt{e}}\sum_{j=-2}^{\infty} c_j n^{-(j+1)/2} K_{-j-1}\left(2\sqrt{n}\right),
$$
where the coefficients $c_j$ are defined by
$$
\exp\left\{\frac{1}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2} = \exp\left\{-\frac{1}{y}-\frac{1}{2}\right\} \sum_{j=-2}^{\infty} c_j y^j.
$$
We can transform this asymptotic series into one in terms of elementary functions by using the known asymptotic expansion for $K_\nu(r)$ as $r \to \infty$ (DLMF reference). In our case we have
$$
K_{-j-1}\left(2\sqrt{n}\right) \sim \frac{\sqrt{\pi}}{2} n^{-1/4} e^{-2\sqrt{n}} \sum_{k=0}^{\infty} \frac{a_k(j+1)}{2^k} n^{-k/2},
$$
where $a_0(j+1) = 1$ and
$$
a_k(j+1) = \frac{1}{k!8^k}\prod_{m=1}^{k} \left(4(j+1)^2-(2m-1)^2\right).
$$
Substituting this into our asymptotic expansion yields
$$
\begin{align}
\frac{2}{\sqrt{e}}\sum_{j=-2}^{\infty} c_j n^{-(j+1)/2} K_{-j-1}\left(2\sqrt{n}\right) &= \sqrt{\frac{\pi}{e}} n^{-1/4} e^{-2\sqrt{n}} \sum_{j=-2}^{\infty} \sum_{k=0}^{\infty} \frac{c_j a_k(j+1)}{2^k} n^{-(j+k+1)/2} \\
&= \sqrt{\frac{\pi}{e}} n^{1/4} e^{-2\sqrt{n}} \sum_{\ell = 0}^{\infty} \left(\sum_{j+k+2=\ell} \frac{c_j a_k(j+1)}{2^k}\right)n^{-\ell/2},
\end{align}
$$
So we conclude that
As $n \to \infty$,
$$
\begin{align}
&\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx \\
&\qquad \sim \sqrt{\frac{\pi}{e}} n^{1/4} e^{-2\sqrt{n}} \sum_{\ell = 0}^{\infty} \left(\sum_{j+k+2=\ell} \frac{c_j a_k(j+1)}{2^k}\right)n^{-\ell/2} \\
&\qquad = \sqrt{\frac{\pi}{e}} n^{1/4} e^{-2\sqrt{n}} \left(1 + \frac{5}{48} n^{-1/2} - \frac{479}{4608} n^{-1} + \frac{15313}{3317760} n^{-3/2} + \cdots \right),
\end{align}
$$
where the indices of the inner sum range over $j=-2,-1,0,1,2,\ldots$ and $k=0,1,2,3,\ldots$.