Following the same process of our previous solutions of this type of problems:
From here we have $$\arcsin^2z=\frac12\sum_{k=1}^\infty\frac{(2z)^{2k}}{k^2{2k \choose k}}$$
Set $z=i\sqrt{\frac{y}{8}}$, we get
$$-\text{arcsinh}^2\left(\sqrt{\frac{y}{8}}\right)=\frac12\sum_{k=1}^\infty\frac{(-1)^{k}y^k}{k^22^k{2k \choose k}}$$
Now multiply both sides by $-\frac{\ln^2 y}{y}$ then integrate from $y=0$ to $1$, we get
\begin{align}
S&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^52^k{2k \choose k}}=\int_0^1\frac{\text{arcsinh}^2\left(\sqrt{\frac{y}{8}}\right)\ln^2y}{y}\ dy,\quad \color{red}{\text{arcsinh}\left(\sqrt{\frac{y}{8}}\right)=x}\\
&=2\int_0^{\frac{\ln2}{2}}x^2\ln^2(8\sinh^2x)\coth x\ dx\\
&\small{=18\ln^22\int_0^{\frac{\ln2}{2}}x^2\coth x\ dx+24\ln2\int_0^{\frac{\ln2}{2}}x^2\ln(\sinh x)\coth x\ dx+8\int_0^{\frac{\ln2}{2}}x^2\ln^2(\sinh x)\coth x\ dx}\tag{1}\\
\end{align}
The first integral is calculated here
$$\int_0^{\frac{\ln2}{2}}x^2 \coth x\ dx=\frac1{16}\zeta(3)-\frac1{24}\ln^32\tag{2}$$
and the second integral is calculated here
$$\small{\int_0^{\frac{\ln2}{2}}x^2\ln(\sinh x)\coth x\ dx=-\frac12\operatorname{Li}_4\left(\frac12\right)+\frac7{16}\zeta(4)-\frac12\ln2\zeta(3)+\frac18\ln^22\zeta(2)+\frac{7}{192}\ln^42}\tag{3}$$
As for the third integral, we calculate it as follows
\begin{align}
I&=\int_0^{\frac{\ln2}{2}}x^2\ln^2(\sinh x)\coth x\ dx,\quad \color{red}{x=\ln y}\\
&=\int_0^{\sqrt{2}}\ln^2y\ln^2\left(\frac{y^2-1}{2y}\right)\left(\frac{y^2+1}{y^2-1}\right)\frac{\ dy}{y},\quad \color{red}{y^2-1=x}\\
&=\frac18\int_0^1\ln^2(1+x)\left(\ln x-\ln2-\frac12\ln(1+x)\right)^2\left(\frac{2}{x}-\frac1{1+x}\right)\ dx\\
&\small{=\frac14\int_0^1\frac{\ln^2(1+x)\ln^2x}{x}+\frac1{16}\int_0^1\frac{\ln^4(1+x)}{x}-\frac14\int_0^1\frac{\ln^3(1+x)\ln x}{x}-\frac12\ln2\int_0^1\frac{\ln^2(1+x)\ln x}{x}\\
+\frac14\ln2\int_0^1\frac{\ln^3(1+x)}{x}+\frac14\ln^22\int_0^1\frac{\ln^2(1+x)}{x}-\frac18\underbrace{\int_0^1\frac{\ln^2(1+x)\ln^2x}{1+x}}_{\large IBP}-\frac1{32}\underbrace{\int_0^1\frac{\ln^4(1+x)}{1+x}}_{\large \frac15\ln^52}\\+\frac18\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{1+x}}_{\large IBP}+\frac14\ln2\underbrace{\int_0^1\frac{\ln^2(1+x)\ln x}{1+x}}_{\large IBP}-\frac18\ln2\underbrace{\int_0^1\frac{\ln^3(1+x)}{1+x}}_{\large\frac14\ln^42}-\frac18\ln^22\underbrace{\int_0^1\frac{\ln^2(1+x)}{1+x}}_{\large\frac13\ln^32}}\\
&\small{=\frac14\underbrace{\int_0^1\frac{\ln^2(1+x)\ln^2x}{x}}_{\Large I_1}+\frac1{32}\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}}_{\Large I_2}-\frac16\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{x}}_{\Large I_3}-\frac12\ln2\underbrace{\int_0^1\frac{\ln^2(1+x)\ln x}{x}}_{\Large I_4}\\+\frac16\ln2\underbrace{\int_0^1\frac{\ln^3(1+x)}{x}}_{\Large I_5}+\frac14\ln^22\underbrace{\int_0^1\frac{\ln^2(1+x)}{x}}_{\Large I_6}-\frac{19}{240}\ln^52}\text{}\tag{4}
\end{align}
(Ignoring $dx$ is intended so no need to edit please)
The result of $I_3$ can be found here
$$\boxed{\small{I_3=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{99}{16}\zeta(5)+3\zeta(2)\zeta(3)-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52}}$$
and the results of $I_4$, $I_5$ and $I_6$ can be found here
$$\boxed{I_4=-4\operatorname{Li_4}\left(\frac12\right)+\frac{15}4\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac{1}{6}\ln^42}$$
$$\boxed{I_5=-6\operatorname{Li}_4\left(\frac12\right)+6\zeta(4)-\frac{21}{4}\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac14\ln^42}$$
$$\boxed{I_6=\frac14\zeta(3)}$$
and now we are left with the remaining integrals $I_1$ and $I_2$ and lets start with the first one.
By using $$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac1{n^2}\right)x^n$$
we get
\begin{align}
I_1&=\int_0^1\frac{\ln^2(1+x)\ln^2x}{x}\ dx=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx\\
&=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{2}{n^3}\right)=4\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-4\operatorname{Li}_5(-1)\\
&=4\left(\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)\right)-4\left(-\frac{15}{16}\zeta(5)\right)\\
&\boxed{I_1=2\zeta(2)\zeta(3)-\frac{29}{8}\zeta(5)}
\end{align}
The sum $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ was nicely done here by M.N.C.E and Cornel Valean.
\begin{align}
I_2&=\int_0^1\frac{\ln^4(1+x)}{x}\ dx\overset{x=\frac{1-y}{y}}{=}\int_{1/2}^1\frac{\ln^4x}{x(1-x)}\ dx\\
&=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\
&=\frac15\ln^52+\sum_{n=1}^\infty \int_{1/2}^1x^{n-1}\ln^4 x\ dx\\
&=\frac15\ln^52+\sum_{n=1}^\infty\left(-\frac{\ln^42}{n2^n}-\frac{4\ln^32}{n^22^n}-\frac{12\ln^22}{n^32^n}-\frac{24\ln2}{n^42^n}-\frac{24}{n^52^n}+\frac{24}{n^5}\right)\\
&=\frac15\ln^52-\ln^52-4\ln^32\operatorname{Li}_2\left(\frac12\right)-12\ln^22\operatorname{Li}_3\left(\frac12\right)-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)+24\zeta(5)\\
&\boxed{I_2=-24\operatorname{Li}_5\left(\frac12\right)-24\ln2\operatorname{Li}_4\left(\frac12\right)+24\zeta(5)+4\ln^3(2)\zeta(2)-\frac{21}2\ln^22\zeta(3)-\frac45\ln^52}
\end{align}
In our calculations, we used the following special values of the dilogarithmic and trilogarithmic functions:
$$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$
$$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$
Plugging the boxed results of $I_1$, $I_2$, $I_3$, $I_4$, $I_5$ and $I_6$ in $(4)$, we get
$$\small{I=\frac54\operatorname{Li}_5\left(\frac12\right)+\frac94\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{19}{16}\zeta(5)-\frac78\ln2\zeta(4)+\frac{95}{64}\ln^22\zeta(3)-\frac{11}{24}\ln^32\zeta(2)+\frac1{240}\ln^52}\tag{5}$$
Finally, by substituting the results of $(2)$, $(3)$, and $(5)$ in $(1)$, we get our closed form:
$$\small{S=10\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{19}{2}\zeta(5)+\frac72\ln2\zeta(4)+\ln^22\zeta(3)-\frac{2}{3}\ln^32\zeta(2)+\frac{19}{120}\ln^52}$$
