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 the length of a world line between two points in spacetime. It is calculated using the metric, which in special relativity is the Minkowski metric:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$
We often define a proper distance instead and this is related to the proper time by $ds^2 = -c^2d\tau^2$. Typically we use the proper time for timelike paths and the proper length for spacelike paths to avoid ending up having to square root a negative number.
The proper time depends on the world line. For any two points $A$ and $B$ there are an infinite number of paths that connect those two points, and those paths will in general have different proper times. However in introductory SR courses we frequently only consider straight lines joining the two points, and in that case for the two points:
$$\begin{align}
A &= (t, x, y, z) \\
B &= (t+\Delta t, x+\Delta x, y+\Delta y, z+\Delta z)
\end{align}$$
the proper time for a straight line joining $A$ and $B$ is simply:
$$ c^2\Delta\tau^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2 \tag{1} $$
Suppose in your example the two event are in the same place i.e. $A$ and $B$ are the same point. Then $\Delta x = \Delta y = \Delta z = 0$ so the proper time calculated using equation (1) is:
$$ \Delta\tau = \Delta t $$
So in this case the proper time is equal to the time interval between the events i.e. the proper time $\Delta \tau$ and the coordinate time $\Delta t$ are the same. But if the two events have a spatial separation the proper time and the coordinate time will be different.
The significance of the proper time is that it is an invariant i.e. all observers in all frames of reference will agree on the value of $\Delta \tau$. However different observers will disagree about the value of $\Delta t$.
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Best Answer
"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"
No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two events along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two events--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).
Also, note that this interval is frame-invariant, which means that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.
Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.