Here I posted a generalized formula for the polylogarithm I had discovered. However, for $x=\frac{1}{2}$, $z=\frac{1}{2}$, $p=1$ wolfram alpha yields a result different than what the double integral is supposed to be equal to. Where is the mistake?
$$ Li_s(z) = \frac{z}{(s- 1)!}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-z}dt. $$
$$ (z\frac{d}{dz})^n Li_s(z) = Li_{s-n}(z). $$
$$ (z\frac{d}{dz})^n Li_s(z) = \sum_{j=0}^n {n \brace j}z^j \frac{d^j}{dz^j}Li_s z$$by the formula from here. $$ = \sum_{j=0}^n {n \brace j}z^j \frac{d^j}{dz^j}\frac{z}{(s-1)!}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-z}dt$$ $$ = \sum_{j=0}^n {n \brace j}z^{j}\frac{1}{(s-1)!}(z\frac{d^j}{dz^j} \int_{0}^{\infty}\frac{t^{s-1}}{e^t-z}dt+j\frac{d^{j-1}}{dz^{j-1}}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-z}dt)$$ $$ = \sum_{j=0}^n {n \brace j}z^{j}\frac{1}{(s-1)!}(z j! \int_{0}^{\infty}\frac{t^{s-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{s-1}}{(e^t-z)^j}dt).$$
Take $$s(n)=c+n.$$
$$ Li_c(z) = \sum_{j=0}^n {n \brace j}z^{j}\frac{1}{(c+n-1)!}(z j! \int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^j}dt)$$
$$ x^{n+c-1}Li_c(z) = \sum_{j=0}^n {n \brace j}z^{j}\frac{x^{n+c-1}}{(c+n-1)!}(z j! \int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^j}dt)$$ $$ \frac{d^{c-1}}{dx^{c-1}}x^{n+c-1}Li_c(z) = \sum_{j=0}^n {n \brace j}z^{j}\frac{x^{n}}{n!}(z j! \int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^j}dt)$$ $$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = \sum_{n=0}^{\infty}\sum_{j=0}^n {n \brace j}z^{j}\frac{x^{n}}{n!}(z j! \int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^j}dt)$$ $$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = \sum_{j=0}^{\infty}\sum_{n=j}^{\infty} {n \brace j}z^{j}\frac{x^{n}}{n!}(z j! \int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^j}dt)$$ $$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = \sum_{j=0}^{\infty} z^{j}(z j! \int_{0}^{\infty}\frac{t^{c-1}}{(e^t-z)^{j+1}}\frac{(e^{tx}-1)^j}{j!}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c-1}}{(e^t-z)^j}\frac{(e^{tx}-1)^j}{j!}dt)$$ by $ \sum_{n=j}^{\infty}{n \brace j}\frac{x^n}{n!}=\frac{(e^x-1)^j}{j!}. $
Introduce the Gamma function: $j!=\int_0^{\infty}h^je^{-h}dh$ $$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = \int_0^{\infty}(\sum_{j=0}^{\infty} z^{j}(z h^j \int_{0}^{\infty}\frac{t^{c-1}}{(e^t-z)^{j+1}}\frac{(e^{tx}-1)^j}{j!}dt+jh^{j-1}\int_{0}^{\infty}\frac{t^{c-1}}{(e^t-z)^j}\frac{(e^{tx}-1)^j}{j!}dt))e^{-h}dh$$
$$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = $$ $$=\int_0^{\infty}\sum_{j=0}^{\infty}z^{j+1}h^je^{-h}\int_0^{\infty}\frac{(e^{tx}-1)^j}{j!}\frac{t^{c-1}}{(e^t-z)^{j+1}}dtdh$$ $$-\int_0^{\infty}\sum_{j=0}^{\infty}z^{j}h^{j-1}e^{-h}\int_0^{\infty}\frac{(e^{tx}-1)^j}{j!}\frac{t^{c-1}}{(e^t-z)^{j}}dtdh$$ $$+\frac{d}{dv}\int_0^{\infty}\sum_{j=0}^{\infty}v^{j+1}h^{j-1}e^{-h}\int_o^{\infty}\frac{(e^{tx}-1)^j}{j!}\frac{t^{c-1}}{(e^t-z)^{j}}dtdh$$ where the derivative w.r.t. $v$ is evaluated at $z$.
$$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = $$ $$\int_0^{\infty}\int_0^{\infty}t^{c-1}\frac{z}{e^t-z}e^{\frac{e^{tx}-1}{e^t-z}zh-h}dtdh$$ $$-\int_0^{\infty}\int_0^{\infty}t^{c-1}\frac{1}{h}e^{\frac{e^{tx}-1}{e^t-z}zh-h}dtdh$$ $$+\frac{d}{dv}\int_0^{\infty}\int_0^{\infty}t^{c-1}\frac{v}{h}e^{\frac{e^{tx}-1}{e^t-z}vh-h}dtdh$$
So in the end,
$$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = \int_0^{\infty}\int_0^{\infty}t^{c-1}(\frac{z}{e^t-z}+\frac{e^{tx}-1}{e^t-z})e^{\frac{e^{tx}-1}{e^t-z}zh-h}dtdh.$$
If this was true, one could get the closed form of Apery's constant. This is because picking $z=\frac{1}{2}, x=\frac{1}{2}$ yields the Gamma function, and the Trilogarithm at $z=\frac{1}{2}$ is known.
Where did I go wrong? As far as I know, it could be convergence issues by introducing the Gamma function. Only that introduction allows for swapping the two nested sums. However, the integral is still convergent. Moreover, this formula works for $x=0$.
I found a slight error towards the end of your derivation.
Your final result should be
$$\frac{\partial^{c-1}}{\partial x^{c-1}}\frac{x^{c-1}}{1-x}\, \text{Li}_c(z)=\int_0^{\infty} \left(\int_0^{\infty} t^{c-1} \left(\frac{z}{e^t-z}+\frac{z\, \left(e^{t x}-1\right)}{e^t-z}\right) e^{\frac{e^{t x}-1}{e^t-z} z h-h} \, dt\right) \, dh\tag{1a}$$ $$=\int_0^{\infty} \left(\int_0^{\infty} t^{c-1}\, \frac{z\, e^{t x}}{e^t-z}\, e^{\frac{e^{t x}-1}{e^t-z} z h-h} \, dt\right) \, dh\tag{1b}$$
which can be rearranged as
$$\frac{\partial^{c-1}}{\partial x^{c-1}}\frac{x^{c-1}}{1-x}\, \text{Li}_c(z)=\int_0^{\infty} \left(\int_0^{\infty } e^{\frac{e^{t x}-1}{e^t-z} z h-h} \, dh\right) \frac{z\, e^{t x}}{e^t-z}\, t^{c-1}\, dt\tag{2a}$$ $$=\int_0^{\infty} \left(\frac{e^t-z}{e^t-z e^{t x}}\right) \frac{z\, e^{t x}}{e^t-z}\, t^{c-1}\, dt\tag{2b}$$ $$=\int_0^{\infty} \frac{z}{e^{t\, (1-x)}-z}\, t^{c-1}\, dt\tag{2c}$$
where the Mellin transform defined in formulas (2b) and (2c) above is valid for $\Re\left(\frac{z\, \left(e^{t x}-1\right)}{z-e^t}\right)>-1$.
For $c=3$ and $z=x=\frac{1}{2}$, formula (2c) above evaluates to the Mellin transform
$$\int_0^{\infty} \frac{\frac{1}{2}}{\left(e^{t\, \left(1-\frac{1}{2}\right)}-\frac{1}{2}\right)}\,t^{3-1} \, dt=\frac{2}{3} \left(21\, \zeta (3)+4 \log ^3(2)-\pi ^2 \log (4)\right)\tag{3}$$
which is equivalent to
$$\frac{\partial^{3-1}}{\partial x^{3-1}}\frac{x^{3-1}}{1-x} \text{Li}_3\left(\frac{1}{2}\right) =\frac{-21 \zeta (3)-4 \log ^3(2)+\pi ^2 \log (4)}{12 (x-1)^3}\tag{4}$$
evaluated at $x=\frac{1}{2}$.
More generally for $x=\frac{1}{2}$ one has
$$\int\limits_0^{\infty} \frac{z}{e^{\left(1-\frac{1}{2}\right) t}-z}\, t^{c-1} \, dt=2^c\, \Gamma(c)\, \text{Li}_c(z)\,,\quad\Re(c)>0\tag{5}$$
and for $x=0$ one has
$$\int\limits_0^{\infty} \frac{z}{e^t-z}\, t^{c-1} \, dt=\Gamma(c)\, \text{Li}_c(z)\,,\quad\Re(c)>0\tag{6}$$
which is equivalent to the integral representation
$$\text{Li}_c(z)=\frac{z}{\Gamma(c)} \int\limits_0^{\infty} \frac{1}{e^t-z}\, t^{c-1} \, dt\,,\quad\Re(c)>0\tag{7}$$
referred to the beginning of the question above.