You write: "From the Spaceship’s point of view, the first signal was sent from Earth when it was 4.8 Spaceship light years away. Thus it would take 4.8 Spaceship years to reach the Spaceship."
Be careful. I, on earth, send out a light signal at the same time (according to me) that the space ship takes off. According to you (on the now-moving spaceship), I sent that signal 10.67 years ago (at time -10.67 by your clock). Also, according to you, I've spent that 10.67 years traveling towards you at speed .8. Therefore, when the light signal was sent, I was not just 4.8 light years away; I was 4.8 plus another (10.67 x .8) light years away --- a total of about 13.33 light years. So the light signal should take a total of 13.33 years to reach you. As it was sent at time -10.67 it should arrive at time 2.67. (I hope I got the arithmetic right!).
The right way to do this, of course, is not to grind through the arithmetic, but to draw the spacetime diagram, which makes everything clear from the start. The black line is the earth, the blue line is the distant planet, the red line is the traveler, and the gold lines are light rays, sent every two years from earth and every two years from the ship. The black numbers are earth times and the red numbers are traveler-times.
Let's call the two points $A$ and $B$, and we'll take them to be a distance $dx$ apart as measured in your frame.
In your frame you measure the other spaceship to take a time $dt$ to get from $A$ to $B$, and since the other spaceship is travelling at a velocity $v$ relative to you that time is given by:
$$ dx = v~dt \tag{1} $$
The proper time, $d\tau$, for the trajectory between the two points is given by the Minkowski metric:
$$ c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 \tag{2} $$
and we will assume all motion is along the $x$ axis so $dy = dz = 0$. Then substituting $dx = vdt$ from (1) we find the proper time is simply:
$$ d\tau = dt\sqrt{1 - v^2/c^2} = \frac{dt}{\gamma} \tag{3} $$
So as I'm sure you suspected just dividing the distance by the velocity does not give you the proper time, it gives you the coordinate time.
It is worth considering what happens in the rest frame of the moving spaceship. In this frame the spaceship is stationary at the origin so it doesn't move in space and $dx'=0$. The point $A$ passes the spaceship at some point $(t' 0)$ and the point $B$ passes at the same point in space but a later time $(t' + dt', 0)$. Then using the metric (2) to calculate the proper time simply gives:
$$ d\tau = dt' \tag{4} $$
So the proper time between the two spacetime points is equal to the time recorded by the spaceship. This is a general result in relativity: the proper time is the elapsed time for the inertial observer travelling between the two points.
And one last point while we're here. If we equate the expressions (3) and (4) for the proper time we get:
$$ dt' = \frac{dt}{\gamma} \tag{5} $$
which is the well known expression for the time dilation of a moving observer.
Best Answer
In the scenario you present, only the observer holding the clock can measure the elapse of its proper time. The moving observer observes his own proper time elapse by looking at the clock that he is holding. Each can calculate the proper time observed by the other, but they observe the proper time of their own clocks only.