I am sharing my proof and would like to see different ways as my solution is pretty long and based on many identities. I am going to start with the right side.
We proved here$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$
Set $a_n=H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}$
We get
$$S_n=\frac1{1-x}\sum_{n=1}^\infty\left[\left(H_n^4-H_{n-1}^4\right)-6\left(H_n^2H_n^{(2)}-H_{n-1}^2H_{n-1}^{(2)}\right)+8\left(H_nH_n^{(3)}-H_{n-1}H_{n-1}^{(3)}\right)\\+3\left(\left(H_n^{(2)}\right)^2-\left(H_{n-1}^{(2)}\right)^2\right)-6\left(H_n^{(4)}-H_{n-1}^{(4)}\right)\right]x^n$$
$$\small{=\frac1{1-x}\left[6\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)x^n+4\sum_{n=1}^\infty\left(\frac{H_n^3}{n}-\frac{3H_nH_n^{(2)}}{n}+\frac{2H_n^{(3)}}{n}-\frac{3H_n^2}{n^2}+\frac3{n^4}\right)x^n\right]}\tag1$$
By Cauchy product we have
$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)x^n\tag2$$.
Also, from the same link above where we proved the following identities:
$$\small{\sum_{n=1}^\infty H_n^3x^n=
\frac{\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)+\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)-3\zeta(3)}{1-x}}$$
$$\small{\sum_{n=1}^\infty H_nH_n^{(2)}x^n=
\frac{\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\frac12\ln x\ln^2(1-x)-\zeta(2)\ln(1-x)-\zeta(3)}{1-x}}$$
$$\sum_{n=1}^\infty\frac{H_{n}^2}{n}x^{n}=\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x)-\frac13\ln^3(1-x)$$
it follows that
$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}-\frac{3H_n^2}{n}+\frac3{n^3}\right)x^n\\=-\frac{\ln^3(1-x)}{1-x}+3\ln(1-x)\operatorname{Li}_2(x)+\ln^3(1-x)$$
Divide both sides by $x$ and integrate to get
$$\require{cancel}\sum_{n=1}^\infty\left(\frac{H_n^3}{n}-\frac{3H_nH_n^{(2)}}{n}+\frac{2H_n^{(3)}}{n}-\frac{3H_n^2}{n^2}+\frac3{n^4}\right)x^n\\=\cancel{-\int\frac{\ln^3(1-x)}{x}\ dx}+\frac14\ln^4(1-x)-\frac32\operatorname{Li}_2^2(x)+\cancel{\int\frac{\ln^3(1-x)}{x}\ dx}$$
Then
$$\sum_{n=1}^\infty\left(\frac{H_n^3}{n}-\frac{3H_nH_n^{(2)}}{n}+\frac{2H_n^{(3)}}{n}-\frac{3H_n^2}{n^2}+\frac3{n^4}\right)x^n=\frac14\ln^4(1-x)-\frac32\operatorname{Li}_2^2(x)+C\tag3$$
where $C=0$ if we set $x=0$.
Plug $(3)$ and $(2)$ in $(1)$ we get
$$S_n=\sum_{n=1}^\infty\left(H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}\right)x^n=\frac{\ln^4(1-x)}{1-x}$$
The key to this proof is how to group up the right terms in $(1)$ so that we get the cancellation of the integral $\int\frac{\ln^3(1-x)}{x}\ dx$. It has a closed form though in case you are curious
$$\small{\int\frac{\ln^3(1-x)}{x}\ dx=\ln^3(1-x)\ln x+3\ln^2(1-x)\operatorname{Li}_2(1-x)-6\ln(1-x)\operatorname{Li}_3(1-x)+6\operatorname{Li}_4(1-x)-6\zeta(4)}$$
which can be found by applying integration by parts couple times together with using the reflection formula of the dilogarithm function $\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)$.
