The notion of "hyperbolic angle" is really a contrived concept that is only used to better define an analogy between the hyperbolic functions as a parameterization of the hyperbola and the trigonometric functions as a parameterization of the circle.
The hyperbolic functions are not defined based on the traditional notion of an angle. They are, in fact, defined based on the area of an intercepted arc on the hyperbola.
To find the area of this intercepted arc using calculus and geometry, subtract the area under the curve from 1 to some x in quadrant I from the triangle created from the line that connects the origin $(0,0)$ with this point $(x,\sqrt{x^2 -1})$:
$\hskip2in$
\begin{align}
A &= \frac{1}{2}x\sqrt{x^2 -1} - \int _1 ^x \sqrt{x^2 -1}dx \\
&= \frac{1}{2}x\sqrt{x^2 -1} - \frac{1}{2}x\sqrt{x^2 -1} + \frac{1}{2}\ln(x + \sqrt{x^2-1}) \\
2A &= \ln(x + \sqrt{x^2-1})\\
e^{2A}&= x + \sqrt{x^2-1}\\
\end{align}
At this point, we can substitute $\sqrt{x^2-1}$ with $y$ for simplicity.
\begin{align}
e^{2A}&= x + y\\
e^{2A}(x-y)&= (x+y)(x-y)\\
e^{2A}(x-y)&= x^2-y^2\\
\end{align}
Because these equations describe a unit hyperbola $x^2 - y^2 = 1$, we can make this crucial substitution and derive the $e^x$ definition.
\begin{align}
e^{2A}(x-y)&= 1\\
e^{2A}&= \frac{1}{x-y}\\
e^{-2A}&= x-y\\
\end{align}
If we add $x+y$ (which, remember, is equal to $e^{2A}$) to this equation, the $y$'s will cancel.
\begin{align}
2x &= e^{-2A} + x + y \\
2x &= e^{-2A} + e^{2A} \\
x &= \frac{e^{-2A} + e^{2A}}{2} \\
\end{align}
We can find $y$ if we calculate $(x+y) - (x-y)$
\begin{align}
(x+y) - (x-y) &= e^{2A} - e^{-2A} \\
2y &= e^{2A} - e^{-2A} \\
y &= \frac{e^{-2A} - e^{2A}}{2}
\end{align}
If we treat these equations as functions themselves, we can now see where this $u = 2A$ definition comes from more evidently. Every $(x,y)$ on the graph $x^2 - y^2 = 1$ can be described in a relationship involving area $A$ times $2$ as proven above. We can substitute $2A$ with $u$ and call our argument $u$ whatever we want henceforth.
Here's an interactive Desmos graph I made to help illustrate the connection. It isn't a "formal proof", but I think its pretty cool.
Best Answer
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:
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 $(\cosh L, \sinh L)$.