OK, I now figured it out myself:
It is not possible to do it in three weighings.
Let's start with some general observations for the first weighing.
For any of the three possible results (one side more heavy, other side more heavy, equilibrium), there have to be at most $9$ possibilities remaining, as the two remaining weighings can at most have $9$ different outcomes. Since there are $24$ possibilities in total, this implies that each result needs to leave at least $6$ possible weight assignments, as the other two can at most cover $18$ possibilities.
Now with this observation, it is easy to weed out all but one possibility for the first weighing:
Obviously an $1:1$ weighing (one object against a single other) cannot give equilibrium, so that can be immediately excluded. A $1:2$ weighing where one of the two lightest objects is the single one will always give that the two balls are more heavy; those already give 12 possibilities, so that obviously doesn't work either. A $1:3$ weighing will necessarily give a heavier weight on the three-object side if the heaviest object is among those three; but if the heaviest object is alone, then you have only one possible result, so at least one result is impossible. So the only viable first weighing is a $2:2$ weighing.
This then gives a first constraint on the weights:
Obviously the most heavy two objects are always heavier than the two lightest object. Also the most heavy and the second-lightest object are together heavier than the two remaining objects. So the only way equilibrium can be achieved is if the most heavy and the lightest object together have the same weight as the two remaining ones.
If this condition is fulfilled,
the $2:2$ measurement gives exactly $8$ remaining possibilities for each of the results, so that is actually perfect.
Now after having identified the correct first measurement, let's look at how to continue:
Let's look at the case where we have equilibrium. Again, we can figure out that any result has to leave at most $3$ possibilities (or otherwise there's no way to distinguish them with a single weighing) and at least $2$ possibilities (or else the other two results would need to cover more than the maximal $6$ possibilities).
For the same reasons as before, we can exclude $1:1$ and $1:3$ weighings. Moreover, the second weighing cannot be another $2:2$ weighing, as the only way we can get more information is if we take on each side one object from each side of the first weighing; but for that configuration we already know that equilibrium cannot be achieved.
So the only weighing that might work as second is a $3:1$ weighing. Let's again consider the case where we have equilibrium.
Obviously, if we have equilibrium, the single ball is one of the two heaviest. Moreover, since the middle two weights are heavier than the most heavy one, we know that the lightest object is among the two objects on the other side of the scale. But this means that for each of the two possible single objects, we can identify which of the two objects is the lightest: If the single object is the most heavy one, the lightest is the one which was on the same side of the scale for the first weighing, otherwise it's the one which was on the other side.
Therefore the only way we can get two possibilities at equilibrium is if both the second-most heavy object weighs the same as the two lightest together, and the most heavy object weighs as much as the second-most heavy and the lightest together.
Together with the condition from the first weighing, this means that if there is a solution at all, it has to be of the form $\{a,2a,3a,4a\}$.
But in that case, the only way the single object is more heavy than the two objects on the other side is if the single object has weight $4a$ and the two objects on the other side are the objects with weights $a$ and $2a$. But that violates the condition that each result needs to leave at least two remaining possibilities.
Therefore there is no combination of four different weights that can be reliably distinguished with only three weighings.
