Is the following Harmonic Number Identity true?
Note: OPs doubts seem reasonable to me, since the identity stated in OPs question is not correct. In the following we are going through @TunkFey's answer and try to put the focus to weak points in his argumentation. But, nevertheless I'd like to emphasize that it was a pleasure to go through his answer which was also instructive and inspiring to me.
Preliminaries
The challenge in Tunk Feys answer was to find a closed expression (at least in terms of Polylogarithms) of \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3} \end{align*} and evaluate it at $x=\frac{1}{2}$.
Raymond Manzoni has nicely demonstrated that \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^2}&=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)\\ &\qquad+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)\tag{1} \end{align*} Tunk Fey took this series as starting point and decided to use the same techniques as Raymond.
(Tunk Fey:) Dividing (1) by $x$ and integrating gives \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3} &=\zeta(3)\ln x+\frac12\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}+\color{blue}{\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}x\ dx}\\&+\operatorname{Li}_4(x)-\color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}\tag{2} \end{align*}
The next step is to simplify \begin{align*} \color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx} \end{align*}
He continues with (question marks emphasis mine)
(Tunk Fey:) Using IBP to evaluate the green integral by setting $u=\operatorname{Li}_3(1-x)$ and $dv=\frac1x\ dx$, we obtain
\begin{align*} \color{green}{\int\frac{\operatorname{Li}_3(1-x)}x\ dx}&=\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}\ dx\qquad x\mapsto1-x\\ &\stackrel{???}{=}\operatorname{Li}_3(1-x)\ln x-\color{blue}{\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}\ dx}.\tag{3} \end{align*}
OP: Why can we add the integrals after the substitution $x \mapsto 1-x$ in the following step? I doubt it since $\int f(x) \ \mathrm{d}x \neq \int f(1-x) \ \mathrm{d}x$ in general.
The substitution $x\mapsto 1-x$ is not valid in Tunk Fey's answer. We can use it e.g. in case of definite integrals by also respecting the limits of the integral. But as OP claimed in general is the substitution not admissible.
We can evalute these expressions using integration by parts to better see the difference. With $u=\frac{\ln x}{1-x}$ and $dv=\operatorname{Li}_2(1-x)dx$ we obtain \begin{align*} \int\frac{\ln x\operatorname{Li}_2(1-x)}{1-x}=\frac{1}{2}\operatorname{Li}_2^2(1-x)+C_1 \end{align*} On the other hand with $u=\frac{\ln (1-x)}{x}$ and $dv=\operatorname{Li}_2(x)dx$ we obtain for the substituted integral (3) \begin{align*} -\int\frac{\ln (1-x)\operatorname{Li}_2(x)}{x}=\frac{1}{2}\operatorname{Li}_2^2(x)+C_2\tag{4} \end{align*} with $C_1, C_2$ constants of integration.
Later on when Tunk Fey needs to determine a constant of integration he evaluates the series $\sum_{n=1}^\infty\frac{\operatorname{H}_nx^n}{n^3}$ at $x=1$. Evaluating the subexpressions at $x=1$ above give
\begin{align*} \left.\frac{1}{2}\operatorname{Li}_2^2(1-x)\right|_{x=1}&=\frac{1}{2}\operatorname{Li}_2^2(0)=0\\ \left.\frac{1}{2}\operatorname{Li}_2^2(x)\right|_{x=1}&=\frac{1}{2}\operatorname{Li}_2^2(1)=\frac{\pi^4}{72} \end{align*} The difference of the integration constants $\Delta=\frac{\pi^4}{72}$ has negative consequences when we look at the final calculation at the end.
With this incorrectly substituted expression (3) Tunk Fey obtains from (2)
\begin{align*} \sum_{n=1}^{\infty}\frac{H_nx^n}{n^3}&=\zeta(3)\ln(x)+\frac{1}{2}\int\frac{\ln x\ln^2(1-x)}{x}\ dx\\ &\qquad+\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{x}\ dx+\operatorname{Li}_4(x)\\ &\qquad-\operatorname{Li}_3(1-x)\ln x+\int\frac{\ln(1-x)\operatorname{Li}_2(x)}{x}\ dx \end{align*}
The next step is to combine $\int\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{x}\ dx$ and $\int\frac{\ln(1-x)\operatorname{Li}_2(x)}{x}\ dx$ with the help of the Euler's reflection formula \begin{align*} \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x) \end{align*}
He obtains \begin{align*} \sum_{n=1}^{\infty}\frac{H_nx^n}{n^3}&=\zeta(3)\ln(x)+\frac{1}{2}\int\frac{\ln x\ln^2(1-x)}{x}\ dx +\operatorname{Li}_4(x)-\operatorname{Li}_3(1-x)\ln x\\ &\qquad+\int\frac{\ln(1-x)\left(\frac{\pi^2}{6}-\ln x\ln(1-x)\right)}{x}\ dx\\ &=\zeta(3)\ln(x)-\frac{1}{2}\int\frac{\ln x\ln^2(1-x)}{x}\ dx+\operatorname{Li}_4(x)-\operatorname{Li}_3(1-x)\ln x\\ &\qquad-\frac{\pi^2}{6}\int\frac{\ln(1-x)}{x}\ dx\\ &=\zeta(3)\ln(x)-\frac{1}{2}\int\frac{\ln x\ln^2(1-x)}{x}\ dx+\operatorname{Li}_4(x)-\operatorname{Li}_3(1-x)\ln x\\ &\qquad-\frac{\pi^2}{6}\operatorname{Li}_2(x)\tag{5} \end{align*}
The next step is to simplify $\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}$. He continues with (question marks emphasis mine)
(Tunk Fey:) Setting $x\mapsto1-x$ and using the identity $H_{n+1}-H_n=\frac1{n+1}$, the red integral becomes \begin{align*} \color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&\stackrel{???}{=}-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\ &=\ldots\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\ &\qquad+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]\tag{6} \end{align*}
The substitution (???) right at the beginning is not justified and not admissible. It will also have some influence when calculating the integration constant at the end.
OP: Why do we omit the constant of integration in the following step? We should add the constant since it will affect the summation.
OPs argument is valid. The RHS of (6) should have a constant of integration included. But note that Tunk Fey introduces in the following step this constant and this aspect is no longer a potential problem.
Continuing with (5) and substituting (6) Tunk Fey obtains \begin{align*} \sum_{n=1}^{\infty}\frac{H_nx^n}{n^3} &=\zeta(3)\ln(x)-\frac{1}{2} \left(\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\right.\\ &\quad\left.+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]\right) +\operatorname{Li}_4(x)-\operatorname{Li}_3(1-x)\ln x-\frac{\pi^2}{6}\operatorname{Li}_2(x)\\ &=\zeta(3)\ln(x)-\frac{1}{4} \ln^2x\ln^2(1-x)+\ln x\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\ln x\operatorname{Li}_3(x)\\ &\quad-\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}+2\operatorname{Li}_4(x) -\operatorname{Li}_3(1-x)\ln x-\frac{\pi^2}{6}\operatorname{Li}_2(x)\tag{7}\\ \end{align*} It follows \begin{align*} \sum_{n=1}^{\infty}\frac{H_nx^n}{n^3} &=\frac{1}{2}\zeta(3)\ln(x)-\frac{1}{8} \ln^2x\ln^2(1-x)+\frac{1}{2}\ln x\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\frac{1}{2}\ln x\operatorname{Li}_3(x)\\ &\qquad+\operatorname{Li}_4(x) -\frac{1}{2}\operatorname{Li}_3(1-x)\ln x-\frac{\pi^2}{12}\operatorname{Li}_2(x)+C\\ \end{align*}
In the last step Tunk Fey determines the integration constant $C$ by setting $x=1$ in (7) and using the known value of \begin{align*} \sum_{n=1}^{\infty}\frac{H_nx^n}{n^3}=\frac{\pi^4}{72} \end{align*} and some special values of Polygamma functions (maybe with some help of Wolfram Alpha).
(Tunk Fey:) Setting $x=1$ to obtain the constant of integration, \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^3}&=\operatorname{Li}_4(1)-\frac{\pi^2}{12}\operatorname{Li}_2(1)+C\\ \frac{\pi^4}{72}&=\frac{\pi^4}{90}-\frac{\pi^4}{72}+C\\ C&=\frac{\pi^4}{60}. \end{align*} Thus \begin{align*} \sum_{n=1}^\infty \frac{H_nx^n}{n^3}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}\tag{8} \end{align*}
OP: Is the identity even true, since putting $x=\dfrac{1}{2}$ gives a [numerically different result][2] than the correct result, as pointed out by the user @Super Abound in the comments of that answer.
Again OPs doubt is reasonable, since due to the inadmissible substitutions in (3) and (6) the integration constant $C=\frac{\pi^4}{60}$ is not correct.
Evaluating the result (8) at $x=\frac{1}{2}$ gives (with some help of Wolfram Alpha) \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&\stackrel{???}{=}-\frac{1}{8}\ln 2\zeta{3}+\frac{1}{24}\left(\ln 2\right)^4+\frac{7\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}1.36998 \end{align*} The RHS giving $1.36998$ is also challenged by user @SuperAbound in the comment section of Tunk Fey's answer.
The correct answer seems to be provided by @Cleo which is \begin{align*} \sum_{n=1}^\infty \frac{H_n}{n^32^n}&=-\frac{1}{8}\ln 2\zeta{3}+\frac{1}{24}\left(\ln 2\right)^4+\frac{\pi^4}{720}+ \operatorname{Li}_4\left(\frac{1}{2}\right)\\ &\stackrel{.}{=}0.55824 \end{align*} The numerical value is also stated in the answer by @MhenniBenghorbal.
The difference of the two answers $\Delta=\frac{\pi^4}{120}$ seems to be a result due to the erroneously performed substitutions.
Epilogue:
The answer of Tunk Fey is full of nice ideas and it's worth that somebody provides a correct proof based upon his calculations. Avoiding the substitutions the main challenge seems to be (see (5)) obtaining an appropriate representation of \begin{align*} \int\frac{\ln x\ln^2(1-x)}{x}\ dx \end{align*} Wolfram Alpha provides a solution for this integral but it looks much too cumbersome.
The sum in question has a closed form in terms of polylogarithms. The proof is complicated, and I don't intend to reproduce it as I derived it some 15 years ago, and polylogs are not a primary interest now. You can always differentiate both sides and use polylog IDs in Lewin.
$$\sum_{k=1}^\infty \frac{y^k}{k^3}H_k=\zeta(4)+2 Li_4(y)-Li_4(1-y)+Li_4(-y/(1-y))+\\ \frac{1}{2} \log(1-y) \Big( \zeta(3) – Li_3(y)+Li_3(1-y)+Li_3(-y/(1-y)) \Big) + \\\frac{1}{12}\log^3(1-y)\log(y) -\frac{1}{24}\log^4(1-y)$$