Without knowing much about the context for this problem, the relationship between the two integrals seems to be pretty direct from integration by parts. For $n, p \geq 1$, writing $u(x) = (\ln x)^n$ and $v(x) = (\ln(1 - zx))^p$, we have
$$\frac{du}{dx} = \frac{n(\ln x)^{n-1}}{x} \qquad \text{and} \qquad \frac{dv}{dx} = -\frac{pz (\ln (1 - zx))^{p-1}}{1 - zx}$$
hence using integration by parts:
\begin{align*}
n\int_0^1 \frac{(\ln x)^{n-1} (\ln(1 - zx))^p}{x} \,dx
&= \int_0^1 \frac{du}{dx} v \,dx \\
&= (uv)|_0^1 - \int_0^1 u \frac{dv}{dx} \,dx \\
&= pz\int_0^1 \frac{(\ln x)^n (\ln (1 - zx))^{p-1}}{1 - zx} \,dx
\end{align*}
where the last equation holds since $u(x)v(x) \to 0$ as $x \to 0$ or $x \to 1$.
This is $(3)$, up to the factor $(-1)^{n+p-1}/n! p!$.
Denote $f(k,j)=\int_0^{\frac{1}{2}} \frac{\log ^j(1-y) \log ^k(y)}{1-y} \, dy$. Then for $j, k>1$ (RHS denotes Beta derivatives)
$$U(k,j):=jf(k,j-1)+kf(j,k-1)=-(-\log(2))^{j+k}+ k \left( \partial_a^{k-1} \partial_b^j B\right) (0,1)$$
Which is direct by IBP, separation, Beta derivatives and reflection $y\to 1-y$: $$\small jf(k,j-1)= -(-\log (2))^{j+k}+ k \int_0^{\frac{1}{2}} \frac{\log ^j(1-y) \log ^{k-1}(y)}{y} \, dy$$
$$\small =-(-\log (2))^{j+k}+ k \left(\int_0^{1}-\int_{\frac{1}{2}}^1 \right) \frac{\log ^j(1-y) \log ^{k-1}(y)}{y} \, dy$$
$$\small =-(-\log(2))^{j+k}+ k \left( \partial_a^{k-1} \partial_b^j B\right) (0,1)-kf(j,k-1)$$
Thus taking $\frac{\binom{n-1}{j-1} \binom{n}{k}}{\binom{n}{j} \binom{n-1}{k-1}}=\frac{j}{k}$ into account yields the important $\color{blue}{formula}$
$$\small \binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)=\frac{\binom{n}{k}\binom{n-1}{m-k} }{-k+m+1}U(k,m+1-k)$$
Now let $y\to\frac{1-x}{2}$ $$I_n=\int_0^{\frac{1}{2}} \frac{\log ^n(2 y) \log ^{n-1}(2 (1-y))}{1-y} \, dy$$
Apply Binomial thm twice, extract $k=0$
$$I_n=\sum _{k=1}^n \sum _{j=0}^{n-1} \binom{n}{k} \binom{n-1}{j} f(k,j) \log ^{2n-j-k-1}(2)+\int_0^{\frac{1}{2}} \frac{\log ^n(2) \log ^{n-1}(2 (1-y))}{1-y} \, dy$$
Take Cauchy product
$$I_n=\sum _{m=1}^{2n-1} \sum _{k+j=m}\binom{n}{k} \binom{n-1}{j} f(k,j) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$
Take care of range of $j,k$
$$\scriptsize I_n=\sum _{m=1}^n \sum _{k=1}^m \binom{n}{k} \binom{n-1}{m-k} f(k,m-k) \log ^{2n-m-1}(2)+ \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \binom{n}{k} \binom{n-1}{m-k} f(k,m-k) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$
Let $k\to m+1-k$, take average
$$\scriptsize I_n=\frac{1}{2} \sum _{m=1}^n \sum _{k=1}^m \left(\binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)\right) \log ^{2n-m-1}(2)+\frac{1}{2} \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \left(\binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)\right) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$
Use the $\color{blue}{formula}$ to simplify
$$\scriptsize I_n=\frac{1}{2} \sum _{m=1}^n \sum _{k=1}^m \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1} U(k,m+1-k)+\frac{1}{2} \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}U(k,m+1-k)+\frac{\log ^{2 n}(2)}{n}$$
Expand $U(k,m+1-k)$
$$ \scriptsize I_n=\frac{1}{2} \left(\sum _{m=1}^n \sum _{k=1}^m +\sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n\right) \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}\left(k \underset{a\to 0}{\text{lim}}\underset{b\to 1}{\text{lim}}\frac{\partial ^{m}B(a,b)}{\partial a^{k-1}\, \partial b^{-k+m+1}}+(-1)^m \log ^{m+1}(2)\right)+\frac{\log ^{2 n}(2)}{n}$$
This is the final expression of $I_n$. According to Lemma $2.3$ in OP's article, all Beta derivatives in this expression lie in the algebra $\mathbb{Q}(\pi^2, \zeta(3), \zeta(5), \zeta(7), \cdots)$, whence after adding up $\log(2)$ terms, $I_n$ lies in the extended $\mathbb{Q}(\log(2), \pi^2, \zeta(3), \zeta(5), \zeta(7), \cdots)$. QED.
Best Answer
Based on Three-sided coin's comments, it seems
$$I_4 = \int_0^1 \frac{\operatorname{Li}_4(x)}{1+x}dx =\ln(2)\zeta(4)+\tfrac34\zeta(2)\zeta(3)-\tfrac{59}{32}\zeta(5) \approx 0.321352\dots$$
(Note: Confirmed as correct by Brevan's answer.) Connecting it to Three-sided coin's other integrals, then
$$I_4 = \ln(2)\zeta(4)-\tfrac12\zeta(2)\zeta(3)-h_1$$ $$I_4 = \ln(2)\zeta(4)+\tfrac14\zeta(2)\zeta(3)-h_2$$
where,
$$h_1=\int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(-x)}}{x}dx = -\tfrac54\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
$$h_2=\int_0^1\frac{\color{blue}{\operatorname{Li}_3(-x)}\ln(1-x)}{x}dx = -\tfrac12\zeta(2)\zeta(3)+\tfrac{59}{32}\zeta(5)$$
which implies,
$$h_1-h_2 =\int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(-x)}{x}dx- \int_0^1\frac{\operatorname{Li}_3(-x)\ln(1-x)}{x}dx =-\tfrac34\zeta(2)\zeta(3)$$
Compare to the similar integrals here that he mentioned,
$$h_3= \int_0^1\frac{\operatorname{Li}_2(x)\,\color{blue}{\operatorname{Li}_2(x)}}{x}dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$
$$h_4 = \int_0^1\frac{\color{blue}{\operatorname{Li}_3(x)}\ln(1-x)}{x}dx =\zeta(2)\zeta(3)-3\zeta(5$$
which has the proven relation,
$$h_3-h_4 = \int_0^1\frac{\operatorname{Li}_2(x)\,\operatorname{Li}_2(x)}{x}dx - \int_0^1\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}dx = \zeta(2)\zeta(3)$$
$$I_n = \int_0^1\frac{\operatorname{Li}_n(x)}{1+x}dx=(-1)^nA(n)+\sum_{k=1}^{n-1}(-1)^{k+1}\eta(k)\zeta(n+1-k)$$ with Dirichlet eta function $\eta(k)$ and $$A(n) = \sum_{k=1}^\infty \frac{H_k}{k^n}(-1)^k$$
is the $n$th "Alternating Euler Sum". However, since,
$$A(n,z) = \sum_{k=1}^\infty \frac{H_k}{k^n}z^k =S_{n-1,2}(z)+\operatorname{Li}_{n+1}(z)$$
for $-1\leq z\leq 1$, then $I_n$ can be expressed by Nielsen polylogs $S_{n,p}(z)$ as I suspected.