Write down what metric this transformation preserves based on animation I made this on desmos:

https://www.desmos.com/calculator/u5qpd135uc

I made it because I wanted to compare and contrast it with the Lorentz boost. 
The transformation should move a point to somewhere on the same green curve depending on the parameter $b$. And the red curves get mapped to each other.

How would I write out the transformation in concrete notation, and how would I determine the metric that this transformation preserves? (i.e. lorentz boost preserves the lorentz metric) Minkowski space.

Maybe it doesn't preserve anything $-$ I don't know at the moment. I'm working on an answer.
 A: I will forget about your parameter $s$, since it just scales $b$. So points are of the form:
$$
(x,t) = \left(\exp\left({\sqrt{n}e^{b}}\right),\exp\left({\sqrt{n}e^{-b}}\right)\right)
$$
If we add a constant $\Delta b$ to $b$, we see that we can write the transformation as:
$$
(x, t)\mapsto (x',t') = \left(x^{\exp(\Delta b)}, t^{\exp(-\Delta b)}\right)
$$
This transformation maps red curves to each other and points stay on the green curves.
"The metric it preserves" is just that points stay on the green curves. These curves are given by 
$$
t=\exp\left(\frac{n}{\log x}\right)
$$
So if $(x,t)$ satisfies the above for some $n$, then also $t'=\exp\left(\frac{n}{\log x'}\right)$. The interesting thing is that $n$ is constant. We can solve for $n$:
$$
n = \log x \cdot\log t
$$
This is a preserved quantity of the transformation. I.e. you might say $\log x \log t$ is a preserved metric of $(x,t)$. 
NOTE: I worked under the assumption $x>1$, $t>1$. In other regions, you may need to adjust the results.
EDIT, according to the discussion in the comments: Let
$$
(z,w) = \frac12(\log x + \log t, \log x - \log t)
$$
I.e. we take log and rotate and scale. Then 
$$
z^2 - w^2 = \log x\log t
$$
so $z^2-w^2$ is invariant. With $k=e^{\Delta b}$, the transformation becomes:
$$
(z',w') = \frac12\left((k+\tfrac1k)z + (k-\tfrac1k)w, (k-\tfrac1k)z + (k+\tfrac1k)w\right)
$$
Finally, if we substitute $1/\sqrt{1-v^2} = (k+k^{-1})/2$, we get:
$$
(z',w') = \frac{1}{\sqrt{1-v^2}}\left(z - vw, w- v z\right)
$$
which is the Lorentz transformation with $c=1$. If we want to introduce $c$ into the original transformation, we could work backwards from here. 
