You are correct about the beautiful chain of eight Wagner-Meerwein rearrangements. The stereochemistry of each of these is given by the fact that this arrangement is suprafacial and proceeds under retention, thus a group on the top of the molecule will stay above the ring plane and a group below will stay below. During the rearrangement process, since each intermediate cation is planar, the rings A to D (from where the hydroxy group was to where the double bond will be) will undergo a ring flip each, preserving the general trans-decalin structure for each pair of rings.
Once the final rearrangement has occurred, we arrive at that position in the molecule which features a cis-decalin system: note how in your drawing of the starting material the final ring E is not aligned with all the others but pointing downward. Assuming that final proton, which is eliminated, would actually migrate, we would generate the final carbocation at the cis-decalin junction. The methyl group between the eliminating proton and carbon A is in the wrong orientation to migrate; the carbon belonging to ring E has a much greater tendency because it is better aligned (anti to the proton). That migration would lead to a 6,5-spiro compound and a trisubstituted double bond. That isn’t bad in itself, but the pathway asked for in the question leads to a trans-decalin system by elimination, which features much less strain than a spiro compound. Additionally, a tetrasubstituted double bond is also more stable.
To arrive at the all-trans-polcyclic system, the proton on the junction of rings D and E must eliminate which it is happy to do. Ring D flips, which now turns the cis-decalin system of DE into a trans-decalin system, ring strain is released and the reaction terminates.