Can observers in different reference frames agree on the time an event took after the fact?

Your setup and intuition are correct:

The station observes different times for the two clocks on the ship. In fact, if we extrapolate and imagine a whole series of synchronized clocks spanning the entire length of the ship, then the station will observe different times on each one of them. It will see the rear clock always ahead of the front clock by a constant time differential. Likewise, it will see the intermediate ones progressively ahead of the front clock, but lagging behind the rear clock. All this is due to relativity of simultaneity. If the station would have a finite dimension along x, then the ship would see a similar situation for clocks at different positions along x on the station. Both the ship and the station observe the other's clocks running slow, yet each of them sees a consistent time sequence in its own frame.

To see how this works let $L$ be the rest length of the ship and $v$ its velocity relative to the station. Denote $beta = frac{v}{c}$ and $gamma = frac{1}{sqrt{1 - beta^2}}$, as usual. Let the station be positioned at $x=0$ in its own frame, and let the ship breach the station's plane at station time $t=0$ and ship time $t'=0$. For future reference let's write the Lorentz transformations between the ship's frame $(x', ct')$ and the station frame $(x, ct)$: $$ x' = gamma (x - beta ct) ct' = gamma (ct - beta x) $$ and $$ x = gamma (x' + beta ct') ct = gamma (ct' + beta x') $$

Your measurements are as follows:

A. The station observes the ship breaching at time $ct=0$, and exiting after its entire length crosses the station plane. Since the station observes the ship length contracted along $x$ by $1/gamma$, the exit time must be $ct = (L/gamma)/beta = L/betagamma$ and measurement A must read $$ cDelta t_A = L/betagamma $$

B. For the clock at the front of the ship, the station sees a time $ct'=0$ when the ship breaches at $ct=0$. When the ship exits, the station will see the front clock at position $x = L/gamma$ and time $ct = L/betagamma$ in station frame. By the Lorentz transformations, this means the ship coordinates of the front clock are $x'= gamma ( L/gamma - beta (L/betagamma) ) = 0$ and $ct' = gamma (L/betagamma - beta (L/gamma) ) = L (1/beta - beta)) = L /betagamma^2$. The front clock time difference as seen station-side is therefore $cDelta t'_F = L/betagamma^2 = cDelta t_A/gamma$. Equivalently, the ship's front clock runs slower than the station clock by a factor of $gamma$, $cDelta t_A = gamma cDelta t'_F$.

Since the time rate is the same for the rear ship clock, the station observes the same time differential on it as on the front clock. Both are time dilated by $gamma$ relative to station time: $$ cDelta t'_R = cDelta t'_F = cDelta t_A/gamma = L/betagamma^2 $$ or $$ cDelta t_A = gamma cDelta t'_R = gamma cDelta t'_F = L/betagamma $$

C. The ship sees its bow breach at $ct'=0$ and its rear exit after the station plane passed its entire length L, so the time between breach and exit on the ship is $$ cDelta t'_C = L/beta $$

D. A reasoning similar to that for measurement B shows that the station time lapse as seen on the ship is $$ cDelta t_D = L/betagamma $$ which is time dilated relative to the ship lapse by a factor of $gamma$: $$ cDelta t'_C = gamma cDelta t_D = L/beta $$

As you can see, the station sees the ship clocks run slowly, while the ship sees the station clock run slowly. Both are right in their own frames and observe consistent sequences of events.

You're driving the ship at speed $3/5$. I'm on the station.

What we agree on:

• When the front of your ship comes through my station, your front clock and my station clock both say 0:00.
• When the back of your ship comes through my station, your rear clock says 20:00 and my station clock says 16:00.

Your explanation:

• The ship is 12 light-hours long (we're not pretending these numbers are realistic). So the trip through the space station, at speed $3/5$, took 20 hours.
• The ship's clocks, which are both accurate, show that the journey started at 0:00 and ended at 20:00.
• But during that time, the station clock advanced only 16 hours, from 0:00 to 16:00. That station clock runs slow!

My explanation:

• The ship is 9.36 light-hours long. So the trip through the space station, at speed $3/5$, took 16 hours.
• When the front of your ship passed through, its clock showed the correct time of 0:00. But your rear clock was showing 7:12
• When the rear of your ship arrived, its clock showed 20:00. Over the course of a 16 hour journey, it had advanced by only 12:48. That rear clock of yours runs slow! (So does your front clock, by the way).

Edited to add: I saw urdv's answer after I had written this. His answer and mine are essentially the same, though if you want to solve problems like this on your own, you'll learn more from his.