Proper time for an accelerating object

In special relativity, you have to choose as frame of reference which is an inertial frame. In this inertial frame, you may consider the movement of any object, whatever this movement is (accelerated or not).

Let the coordinates of the moving object, relatively to an inertial frame $F$, be $x$ and $t$. We can consider an other initial frame $F'$, which coordinates of the moving object, relatively to $F'$, are $x'$ and $t'$

The heart of special relativity is that exists an invariant which is $c^2 dt^2 - \vec dx^2 = ds^2$. This means that : $c^2 dt^2 - \vec {dx}^2 =ds^2= ds'^2 = c^2 dt'^2 - \vec {dx'}^2$. All inertial frames, when looking at the moving object, agree on the same value $ds^2$

Now, at some instant $t_0$, you may always consider a inertial frame $F'(t_0)$ which has, at this instant, the same speed as the moving object, relatively to $F$. Of course, you will have a different inertial frame $F'(t)$ for each instant. However, the key point is, that the instantaneous speed of the moving object relatively to $F'(t)$ is zero, that is, you have $dx' =0$, so you may write : $ds^2 = c^2 dt^2 - \vec {dx}^2 = ds'(t)^2=c^2dt'^2$

The time $t'$ defined in this manner is called the proper time of the moving object, and is noted $\tau$ ($c^2 dt^2 - \vec {dx}^2 = c^2d\tau^2$). It represents the time elapsed for a clock moving with the moving object.

With your problem, note that if you take the parametrization :

$$\left\{ \begin{array}{l l} ct= b ~sh (\frac{c \tau}{b}) \tag{1}\\ x= b ~ch (\frac{c \tau}{b}) \end{array}\right.$$ you will find, with a little algebra, that, first, $x(t) = \sqrt{(b^2)+((ct)^2)}$, and secondly, that $c^2 dt^2 - \vec {dx}^2 = c^2d\tau^2$ (We suppose here $dy=dz=0$).

So, $\tau$ is the proper time, that you are looking for, and you may find a expression of $\tau$ relatively to $t$, by inversing the first equation of the parametrization $(1)$ :

$$ \tau = \frac{b}{c}~Argsh (\frac{c t}{b}) \tag{2}$$

The proper time is well defined in SR, SR + acceleration and indeed GR, and is invarient in all three. In this case the invarience of the proper time means you can just use the elapsed time for the observer at rest.

The equation you've been given is a thinly disguised version of the relativistic rocket equation:

$$ d(t) = \frac{c^2}{a} \left(\sqrt{1 + \left(\frac{at}{c}\right)^2} - 1 \right) $$

($c$ is the speed of light here - it isn't clear if $c$ means the speed of light or just a constant in the equation you've been given). The derivation of the equation for the relativistic rocket is given in Gravitation by Misner, Thorne and Wheeler, chapter 6.