LPDDR does many, many things to reduce power consumption (the current version even does DVFS, which I think is a first for main memory; graphics cards have been doing interface DFS for a long time, but a high end graphics card will burn around 30-70 W in the memory chips, and uses extremely high speed PHYs running at around 20 GBit/s per pin in the current generation).
The only thing that really changes between DIMM and soldered on is a somewhat increased trace length. That accounts only for a very small increase in power.
PHY is only part of the DRAM chip, and trace length is only a part of what determines energy per bit, so increasing trace length by x % wouldn't increase overall power by x %, but a fraction of that. Thinking about it, there is a bigger problem than just energy/bit scaling; since LPDDR switches dynamically between high-speed and lower speed modes, the high speed mode is likely not used a lot. However, the low speed modes rely on being able to turn the termination off, which saves significant amounts of power. A socket may make that difficult to achieve, and if you have to run with ODT in the, I'm assuming, mostly-used low power mode, you might end up with a significant increase in power.
Of course there are multiple factors at play, but that observation doesn’t justify the conclusion that any particular factor makes only a ‘very small l’ contribution. Do you have even a back-of-an-envelope calculation to support this claim? The knock-on effects of increased trace length can be quite complex.
The only thing that really changes between DIMM and soldered on is a somewhat increased trace length. That accounts only for a very small increase in power.