Circular angles vs. hyperbolic angles

You can, but it requires a little ingenuity.

It should be obvious that if you try to define the length of the arc via tge usual Euclidean measure

$L=\int\sqrt{dy^2+dx^2}=\int{\sqrt{(dy/dx)^2+1}}dx$

you will get nowhere near where you want to go.

Instead define an alternative, non-Euclidean metric:

$L=\int\sqrt{dy^2-dx^2}=\int{\sqrt{(dy/dx)^2-1}}dx$

Let us see what this metric gives for the hyperbola defined by $x=\cosh t, y=\sinh t$:

$dy^2-dx^2=(\cosh^2t-\sinh^2t)dt^2=dt^2$

So we may render the differential length as $dt$ and then integrating from $t=0$ to$t=\alpha$ gives, indeed, $\alpha$ "hyperbolic radians".

You may want to look up how proper time and distance intervals are defined in Einstein's Theory of Special Relativity:

In special relativity, however, the interweaving of spatial and temporal coordinates generates the concept of an invariant interval, denoted as ${\displaystyle \Delta s^{2}}$:

${\displaystyle \Delta s^{2}\;{\overset {def}{=}}\;c^{2}\Delta t^{2}-(\Delta x^{2}+\Delta y^{2}+\Delta z^{2})}$[note 6]

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Given the comments, the question is posed: Can we derive the hyperbolic sine and cosine a priori from the arc length definition used above?

Our problem may be stated as follows: How do $x$ and $y$ vary with arc length, defined by the metric

$ds^2=dy^2-dx^2$

from an initial point $(1,0)$ along the curve

$x^2-y^2=1$?

Begin by differentiating the equation of the hyperbola:

$2xdx-2ydy =0$

$\dfrac{dx}{dy}=\dfrac{y}{x}$

Substituting this into the metric definition and isolating the derivative leads to

$(\dfrac{ds}{dy})^2=\dfrac{1}{y^2+1}$

$(\dfrac{dy}{ds})^2={y^2+1}$

To solve this last equation we can differentiate it. Using the Chain Rule:

$2\dfrac{dy}{ds}\dfrac{d^2y}{ds^2}=2y\dfrac{dy}{ds}$

We cannot have $dy/ds=y^2+1=0$, so:

$\dfrac{d^2y}{ds^2}=y$

and by the usual methodology for linear differential equations with constant coefficients

$y=Ae^s+Be^{-s}$

We need two initial conditions. First from the problem statement we must have $y=0$ at $s=0$. Second, $(\dfrac{dy}{ds})^2={y^2+1}$ implies $dy/ds=1$ at $y=0$ which in turn was just matched with $s=0$ (positive $s$ is taken to be positive $y$, which is basically just a sign convention). From these conditions is obtained

$\color{blue}{y=\dfrac{e^s-e^{-s}}{2}\overset{def}{=}\sinh s}$

And then it's all algebra, using the fact that $(e^s+e^{-s})^2-(e^s-e^{-s})^2=4$ and the curve is confined to positive $x$ by construction:

$\color{blue}{x=\sqrt{1+y^2}=\dfrac{e^s+e^{-s}}{2}\overset{def}{=}\cosh s}$

So, the result of a displacement of length $L$ from $(1,0)$ along $x^2-y^2=1$ may indeed be rendered as $(\cosh L, \sinh L)$.

The red region in your diagram can be parameterized as $x=\rho\cosh\phi,\,y=\rho\sinh\phi$ for $\rho\in[0,\,1],\,\phi\in[0,\,a]$. We've thus related Cartesian coordinates to another coordinate system, with Jacobian matrix$$J=\left(\begin{array}{cc} x_{\rho} & x_{\phi}\\ y_{\rho} & y_{\phi} \end{array}\right)=\left(\begin{array}{cc} \cosh\phi & \rho\sinh\phi\\ \sinh\phi & \rho\cosh\phi \end{array}\right),$$of determinant $\rho$, so $dxdy=\rho d\rho d\phi$. So the red area is$$\int_0^1\rho d\rho\int_0^a d\phi=\frac12a,$$but you already knew that. Meanwhile, the conditions$$dx=\cosh\phi d\rho+\rho\sinh\phi d\phi,\,dy=\sinh\phi d\rho+\rho\cosh\phi d\phi$$simplify on the $\rho=1$ arc to$$dx=\sinh\phi d\phi,\,dy=\cosh\phi d\phi\implies ds=\sqrt{\cosh 2\phi}d\phi.$$The Euclidean arc length is therefore$$\int_0^a\sqrt{\cosh 2\phi}d\phi=\int_0^a\sqrt{1+2\sinh^2\phi}d\phi=-2iE\bigg(\frac{ia}{2}\bigg|2\bigg)$$in terms of elliptic integrals. By contrast, $ds=\sqrt{dy^2-dx^2}$ gives the desired result with a Lorentzian pseudometric, as @OscarLanzi explained.