Let's call the time interval of washing the windows in the train's reference frame $\Delta t$ and the time interval in the stations reference frame $\Delta t'$. As you alude to, the observers at the train station will measure this time interval to be longer than those on the train, specifically $\Delta t' = \Delta t \gamma$ where $\gamma > 1$ when the relative motion is not $0$.
This essentially means that the time interval that would pass from the person on the train started washing his windows till he ended would seem longer from the perspective of the observer on the station. The closer to the speed of light he is moving at, the slower he would seem to be moving.
As a more illustrative example, consider the following:
Lightning strikes the end of a moving train car. The train car has a length of L. To an observer located at the other end of the train car, the flash will reach him at the time,
$$ t = \frac{L}{c} $$
where $c$ is the speed of light, which is always the same for all observers in relativity. For an observer on the ground though, the train will have moved slightly between the two events, meaning that the light flash will have traveled a slightly different distance. This distance could be longer or shorter depending on the direction of the trains movement. Let's call this change in distance $\Delta L$. We then get the time as measured in the ground reference frame
$$t' = \frac{L+\Delta L}{c}$$
which illustrates that, if the speed of light is the same for all observers, the time it takes light to travel the length of the train is different for the two observers. This is analogous to your example of washing windows, except instead of the train moving slow compared to $c$ and the movement inside the train (lightning flash) being close to $c$, the situation is reversed (train speed close to $c$, speed of washing hands not close to $c$).
The important thing to remember here is that special relativity only works on inertial reference frames (i.e. no acceleration), so the two observers could never meet to compare clocks, except for possibly one single occasion when they pass each other (which is generally considered the point where $t = t' = 0$). This means there is no "true time" in special relativity. They are both correct in their analysis, even though they would disagree on the time interval if they could meet which, again, they can't unless one of them accelerates. (In the case of acceleration, general relativity would be required to analyze the scenario, which I am not familiar with... although I know time is still funky. There is no absolute time).
It's always tricky to gain physical understanding using time dilation/ length contraction without talking about relativity of simultaneity. It is actually this last phenomena that truly helps preserve the speed of light after switching inertial frames.
Mathematically, this is all encapsulated in the Lorentz transformation. Say the star uses the reference frame $(x,t)$ and the ship uses $(x',t')$, and it is moving at velocity $v$ with respect to the star. The transformation now reads (with $\gamma = 1/\sqrt{1-v^2/c^2}$):
$$
x' = \gamma(x-vt)
$$
$$
t' = \gamma(-vx/c^2+t)
$$
(sanity check: $x'=0$ corresponds to the wordline $x=vt$) or conversely:
$$
x = \gamma(x'+vt')
$$
$$
t = \gamma(vx'/c^2+t')
$$
The fact that $t'$ depends on $x$ as well is the matheatical translation of relativity of simultaneity, while the extra $\gamma$ factors translate time dilation.
A light path has the world line $x=\pm ct$, plugging in the Lorentz transformation, this becomes $x'=\pm ct'$, so you recover your result. You'll notice that in the end, time dilation did not play an important role at all.
You can visualize these abstract calculations with a nice space-time diagram, but MinutePhysics already does a very good job at illustrating it in its SR playlist. Hope this helps and tell me if you find some mistakes.
Best Answer
Your last statement got it right. One of the postulates of special relativity is the principle of relativity, stating that the laws of physics take the same form in all inertial frames. This statement includes the Lorrentz transformations, and so observers in your frame would measure that earth clocks would be slow by a factor of $ \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $. Similarly, observers on earth would assert the same about your clocks.
What's happening here is the relativity of phase. Two events judged to be simultaneous in one frame will not be so in another inertial frame. So, say, you observe that earth 's clocks have elapsed $t$ seconds, when yours have elapsed $t'$ seconds. In your frame, these events are simultaneous, but in the earth frame, they aren't. As long as you remain in this frame of reference, that is, you do not turn around, this state will continue.
More specifically, describing what an observer "sees" is not the same as what they "measure.". Instead, one must consider the relativistic Doppler effect. However, in the case where our rocket is simple receding from Earth, the result is simple, and the same as our above analysis - the clocks on Earth appear slow