$\DeclareMathOperator{\sech}{sech}$
$\DeclareMathOperator{\Si}{Si}$
$\DeclareMathOperator{\erf}{erf}$

Below is an attempt to prove your equation for $b=1$.

Take $s=\sigma+ti$ with $\sigma,t \in \mathbb{R}$ and $\eta(s) = \left(1-2^{1-s}\right)\zeta(s)$.

Note that:

$$\eta(s) = \frac12 \int_{-\infty}^{+\infty} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^s} du \qquad s \in \mathbb{C} \tag{1}$$

The function to prove is:

$$\eta(\sigma+ti)=\frac{1}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^2)-1\right)\cos\left(x\,\sqrt[4]{-1}\sqrt{t}\right)\,\cos(xy)\,dx\,dy +1\tag{2}$$

Let's start with the real part ($t=0$) which gives:

$$\eta(\sigma)=\frac{2}{\pi}\int_{0}^{\infty}\left(\eta(\sigma+y^2)-1\right)\,\int_{0}^{\infty}\cos(xy)\,dx\,dy +1\tag{3}$$

Simplifiying the integral over $x$ and expanding the domain over $y$ gives:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{\pi}\int_{-v}^{v}\left(\eta(\sigma+y^2)-1\right)\,\frac{\sin(vy)}{y}\,dy+1 \tag{4}$$

Let's remove the $-1$ by observing that $\frac{1}{\pi}\int_{-v}^{v}-\frac{\sin(vy)}{y} dy = -\frac{2\Si(v^2)}{\pi}$ where Si = Sine Integral:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{\pi}\int_{-v}^{v}\eta(\sigma+y^2)\,\frac{\sin(vy)}{y}\,dy-\frac{2\Si(v^2)}{\pi} +1 \tag{5}$$

Now inject integral (1) for $\eta(s)$:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{2\pi}\int_{-v}^{v} \int_{-v}^{v}\frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma+y^2}}\,\frac{\sin(vy)}{y}\,dy\,du -\frac{2\Si(v^2)}{\pi} +1 \tag{6}$$

Which allows the $y^2$ to move to the right:

$$\eta(\sigma)=\lim_{v\to\infty} \frac{1}{2\pi}\int_{-v}^{v} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\int_{-v}^{v}\frac{1}{\left(\frac12+ui\right)^{y^2}}\frac{\sin(vy)}{y}\,dy\,du -\frac{2\Si(v^2)}{\pi} +1 \tag{7}$$

The $y$-integral now nicely evaluates as: $\pi \erf\left(\frac{v}{2\log(1/2+ui)} \right)$, with erf = error function, which is always $\pi$ independent of $u$ when $v \rightarrow \infty$.

With $\displaystyle \lim_{v\to\infty}\frac{2\Si(v^2)}{\pi} = 1$ we then obtain the desired result:

$$\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\pi\,du -1 +1 = \eta(\sigma) \tag{8}$$

Side comment: equation (7) is also valid for $\sigma \in \mathbb{C}$.

For $t \ne 0$, the situation turns out to be a bit more complicated. Starting from equation (7):

$$\eta(\sigma,t)=\lim_{v\to\infty} \frac{1}{2\pi}\int_{-v}^{v} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\int_{-v}^{v}\frac{1}{\left(\frac12+ui\right)^{y^2+ti}}\frac{\sin(vy)}{y}\,dy\,du -\frac{2\Si(v^2)}{\pi} +1 \tag{9}$$

which already gives a partial 'separation' between the real ($\sigma$) and imaginary ($t$) parts. Using the following relation for $\Re(z) > 0, a \in \mathbb{C}$ that I found numerically (hard proof required, asked here):

$$\lim_{v\to\infty} \int_{-v}^{v}\frac{1}{z^{y^2}}\frac{\sin(v\,(\sqrt{a}+y))}{\sqrt{a}+y}\,dy = \frac{\pi}{z^{a}} \tag{10}$$

and with $z=\frac12+ui, a = ti$ the final integral becomes:

$$\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\sech(\pi u)}{\left(\frac12+ui\right)^{\sigma}}\,\frac{\pi}{\left(\frac12+ui\right)^{ti}} \,du = \eta(\sigma,t) \tag{11}$$

which is the desired outcome (note $\sqrt{ti}=\sqrt[4]{-1}\sqrt{t}$).

**Couple of observations:**

- The proof shows that $\sigma$ and $ti$ could also be 'swapped' as follows:

$$\eta (s)-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(ti+y^2)-1\right)\cos(\sqrt{\sigma}\,x)\cos(x y)\,dy\,dx \tag{12}$$

or even stretch it to:

$$\eta (s)-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(y^2)-1\right)\cos(\sqrt{s}\,x)\cos(x y)\,dy\,dx \tag{12}$$

- Other combinations than $y^2$ and $\sqrt{ti}$ are allowed, e.g.:

$$\eta (s)-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^4)-1\right)\cos(\sqrt[4]{ti}\,x)\cos(x y)\,dy\,dx \tag{13}$$

- I believe the proof still works for $b \ne 1$ (the u-integral 'contracts' back to its original form):

$$\eta (s)^b-1=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\left(\eta(\sigma+y^2)^b-1\right)\cos(\sqrt{ti}\,x)\cos(x y)\,dy\,dx \tag{14}$$

- The mechanism seems to work for a broader class of entire functions like $(s-1)\zeta(s), \eta(s), \beta(s), \frac{1}{\Gamma(s)}, \sin(s)$ (i.e. it is not specific for Dirichlet series).