Or would there be a fifth geometry as well?
Short answer: no
TL;DR answer: No, because there can only be pentagons next to pentagons next to more pentagons, or a pentagon surrounded by trapezoids, or a rhombus surrounded by rhombuses surrounded by rhombuses, or three rhombuses surrounded by six trapezoids surrounded by three rhombuses.
To answer your question, we must provide a few facts about these constructions:
(1) The only faces we are allowed to have on these polyhedra are pentagons, trapezoids, and rhombuses
(2) The construction must be convex at all points, barring minimal "fudging".
Two other limiting constraints due to how the pieces work are as follows:
(3) A vertex that is on an acute angle must have four acute angles meeting up.
(4) A vertex that is on an obtuse angle must have three obtuse angles meeting up.
(5) The center caps of the puzzle are completely arbitrary. For this demonstration, I will be referring only to the rough shape of the face and not the center cap.
(6) A rhombic face has two acute angles opposite each other, and two obtuse angles opposite each other.
(7) A trapezoidal face has two obtuse angles next to each other, and two acute angles next to each other.
(8) A pentagonal face has five obtuse angles next to each other.
If we have a pentagon adjacent to another pentagon, then we must fill each vertex with a three obtuse-angled polygons. The only polygon that fits next to two adjacent pentagons is a third pentagon, since there are three obtuse vertices that we must join. the next polygon we place adjacent to the other two of the first three must also be a pentagon, since we run into the same problem as when we placed the third pentagon. This process continues until we have placed twelve pentagons. This forms the Bermuda Megaminx.
Since there are no other possible constructions involving two adjacent pentagons, let us consider a pentagon adjacent to a trapezoid. There is only one orientation that the trapezoid can be adjacent to the pentagon since there is only one face that has two adjacent obtuse angles. Placing this down, the next polygon adjacent to the first two must also be a trapezoid, since it has an acute angle next to the trapezoid, followed by the two obtuse angles adjoining the pentagonal face. We see that the next spot we can fill has the same property, so we repeat this five times, forming a hemisphere constructed with a pentagon surrounded by five trapezoids. From here, there is no other polygon you can place next to any one of the five trapezoids besides another trapezoid. The construction continues trivially until we have five of the six remaining spots filled, which leaves a pentagon filling the last spot. This forms the Bermuda Jumblix.
We cannot have a rhombus adjacent to a pentagon since this would mean having an acute angle next to an obtuse angle.
That leads us to our next problem, a construction with no pentagons at all.
We will start this construction by choosing a rhombus as our first piece. A rhombus has two acute adjacent to one obtuse angle, so we will put a second rhombus next to the first one making sure that the acute angles are only touching acute, and the obtuse angles are only touching obtuse. This leaves an unfilled obtuse angle that is next to two acute angles, which means that only a rhombus can fill that spot.
We are left with three obtuse three-fold vertices with three acute four-fold vertices between them. There are only two ways we can fit our given polygons into this arrangement, one with trapezoids and the other with rhombuses. I will start with rhombuses.
We place a rhombus down next to one of the three rhombuses that are already there such that the obtuse and acute vertices meet with their respective vertices. There is a gap in the four-fold acute vertex that can only be filled with another rhombus. What is left is an obtuse vertex that is surrounded by two acute vertices, so we can only place another rhombus in that spot. The construction continues trivially until we have filled up the six surrounding faces of the first three rhombuses we placed. the holes that we are left with all have acute angles followed by obtuse angles, which can only be filled with three rhombuses. This construction forms the Bermuda Comet.
The final construction involves going back to our three initial rhombuses, but instead we place trapezoids where we originally placed rhombuses. Recall that we had an obtuse angle next to an acute angle which was next to an obtuse angle. we place a trapezoid on its end such that its obtuse angle meets with the obtuse angle that is already there, and its acute angle meets with the acute angle that is already there. The gap formed next to this trapezoid has an obtuse angle, followed by two acute angles adjacent to the trapezoid we just placed. The only shape that fits that spot is another trapezoid. We keep following the construction, and we get an acute angle, followed by two obtuse angles from the previous trapezoid. This construction continues three more times in this fashion until we are left with six trapezoids surrounding the three initially placed rhombuses. the holes in the polyhedron left all have an obtuse angle surrounded by two acute angles. The only pieces that fit these spots are three rhombuses. This construction forms the Bermuda Rhomdo, the puzzle featured in this video.
As you can see, there are no more constructions that could exist because there are simply no other ways to fit together these three polygons, so the answer to your question is that no, there are no other puzzles you can form by simply making another core.
There may be more puzzles like this that have gaps in the puzzle, so to speak. I am not counting these, and they are outside the scope of these constructions.