I usually don't answer my own questions, but I literally just derived the solution and I think it's beautiful.
So we know from the link in the OP that:
$$\zeta(s,a)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{z^{s-1} dz}{e^{a z} (1-e^{-z})}$$
Which makes our expression:
$$\zeta \left(\frac12, \frac{t}{2}\right)-\zeta \left(\frac12, \frac{t+1}{2}\right)=\frac{1}{\sqrt{\pi}} \int_0^\infty \frac{e^{- t z/2}(1-e^{-z/2}) dz}{(1-e^{-z}) \sqrt{z}}=$$
$$=\frac{2}{\sqrt{\pi}} \int_0^\infty \frac{e^{- t u^2/2}(1-e^{-u^2/2}) du}{1-e^{-u^2}}$$
Now we can see that it's very easy to take integral over $t$ (integration by parts):
$$\int_0^1 \sin \pi t~ e^{- t u^2/2} dt=\frac{4 \pi}{4 \pi^2 +u^4} (1+e^{-u^2/2})$$
Now we substitute this into the second integral to get (this is the beautiful part):
$$8 \sqrt{\pi} \int_0^\infty \frac{(1+e^{-u^2/2})(1-e^{-u^2/2}) du}{(1-e^{-u^2})(4 \pi^2 +u^4)}=8 \sqrt{\pi}\int_0^\infty \frac{du}{4 \pi^2 +u^4}$$
After a change of variables we have:
$$\frac{2 \sqrt{2}}{\pi} \int_0^\infty \frac{dv}{1 +v^4}=\frac{2 \sqrt{2}}{\pi} \frac{\pi}{2 \sqrt{2}}=1$$
Just as it was supposed to be.
God, Mathematics is simply perfect sometimes.
Appendix
Trying to prove in a simple way that the integral formula for $\zeta(1/2,a)$ equals the series definition.
$$ \int_0^\infty \frac{z^{-1/2} dz}{e^{a z} (1-e^{-z})}=2 \int_0^\infty \frac{e^{-a u^2}du}{ 1-e^{-u^2}}=2 \sum_{n=0}^\infty \int_0^\infty e^{-(a+n) u^2} du=$$
$$=2 \sum_{n=0}^\infty \frac{1}{\sqrt{a+n}} \int_0^\infty e^{-w^2} dw= \sqrt{\pi} \sum_{n=0}^\infty \frac{1}{\sqrt{a+n}} $$
Formally, this exactly fits the series definition, but it doesn't converge (it's alright, as the function for $s=1/2$ is defined by analytic continuation).
On the other hand, for the particular function in my case, the proof works well, as the series inside the integral becomes alternating due to a factor $(1-e^{-u^2/2})$ in the numerator, so everyting converges.
I would say that all of this constitutes a nice real proof of the Fresnel integrals, especially since the Poisson integral also has a few real proofs.
Though I'm definitely not claiming this is a new result. It was new for me, but a two second google search found me this paper https://www.jstor.org/stable/2320230, and I'm sure there's plenty more.
Best Answer
Starting off after your first substitution, notice that $$\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}, \text{ for } x \in(-1,1)$$ Since the domain of $x,y$ is $(0,1)$, we can write $$\frac{e^t}{(1+e^t)^2}=-\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}$$ In addition, you made a slight error when calculating $dt$ I suppose. It should be $y$ not $y^2$ in the denominator. $$\int_0^1 \int_0^{\ln{y}} \frac{t^4}{y}\left( -\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}\right) \; dt \; dy$$ Because the summation converges, we can interchange the summation and integral sign from Fubini's theorem: \begin{align} k &= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{1}{y} \int_0^{\ln{y}} t^4 e^{tn}\; dt \; dy \\ &\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{y^{n-1} \left(n^4 \ln^4{y}-4n^3\ln^3{y}+12n^2\ln^2{y}-24n\ln{y}+25\right)}{n^5} \; dy \\ &\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \cdot \frac{120}{n^6} \\ &= 120\sum_{n=1}^{\infty} \frac{{(-1)}^{n+1}}{n^5} \\ &= \boxed{\frac{225}{2}\zeta(5)} \\ \end{align}