It is fairly straightforward to see that the hyperbola $xy=1$ is simply the hyperbola $x^2-y^2=2$ rotated by $\pi/4$. All we do is apply the corresponding rotation matrix to the vector $(x_0,y_0)=(\sqrt{2-y^2},y)$ to find that $$\begin{bmatrix} \cos(\pi/4) & -\sin(\pi/4) \\ \sin(\pi/4) & \cos(\pi/4) \end{bmatrix}\begin{bmatrix} \sqrt{2-y_0^2} \\ y_0 \end{bmatrix}=\cos(\pi/4)\begin{bmatrix} \sqrt{2-y_0^2}-y_0 \\ \sqrt{2-y_0^2}+y_0 \end{bmatrix}=\begin{bmatrix}x_1 \\ y_1 \end{bmatrix}$$ It is not difficult to see that $x_1y_1=1$, i.e. that they do indeed lie on the desired hyperbola and we can use the $y$ coordinate of our original hyperbola (which we've been calling $y_0$) to find the 1-to-1 correspondence between points. Before, however, we could nicely parameterize our hyperbola using the hyperbolic angle, $\phi_0$, where $x_0=2\cosh(\phi_0)$ and $y_0=2\sinh(\phi_0)$.
My first question is, do we really have to conjugate by a rotation to find the relation between the hyperbolic trig functions and the hyperbola $x_1y_1=1$? That seems awfully labor-intensive, not to mention unsatisfying. Maybe I'm just being a bit dim but I'm not seeing another way.
My second question, assuming that there is no neat way to parameterize our new hyperbola using the traditional hyperbolic trig functions, is, can we still nicely parameterize our new hyperbola $x_1y_1=1$ using some other 'hyperbolic trigonometric functions of the second kind'? The regular hyperbolic trig functions work nicely for parameterization of the original hyperbola because, first and foremost, they satisfy the identity $x^2-y^2=1$ but I do not know of a similar pair of functions which satisfy $f(\phi_1)g(\phi_1)=1 $ $ \forall \phi_1$ and which have the other nice properties (parity under differentiation, double angle formulas, etc) we'd like them to have? Clearly their MacLaurin Series will not converge but that perhaps they can be expressed in terms of elementary functions like the regular hyperbolic functions. Any and all insights are welcome!
The hyperbolic functions are linearly related to the exponentials.
$$\cosh t=\frac{e^t+e^{-t}}2,\sinh t=\frac{e^t-e^{-t}}2.$$
Obviously,
$$e^te^{-t}=1,$$
which is probably the alternative parameterization you are thinking of.
[This hints to use complex numbers to deal with arbitrary rotations,
$$z=(e^t+ie^{-t})e^{i\theta}=e^{t+i\theta}+ie^{-t+i\theta},$$
but that doesn't seem to further simplify.]
You can also stay in the algebraic domain and simply use
$$t\frac1t=1.$$