1. Fill a grid with all 6s, then topple it.
2. Subtract the result from a fresh grid with all 6s, then topple it.
So effectively it's computing 'all 6s' - 'all 6s' to get an additive identity. But I'm not entirely sure how to show this always leads to a 'recurrent' sandpile.
EDIT: One possible route: The 'all 3s' sandpile is reachable from any sandpile via a sequence of 'add 1' operations, including from its own successors. Thus (a) it is a 'recurrent' sandpile, (b) adding any sandpile to the 'all 3s' sandpile will create another 'recurrent' sandpile, and (c) all 'recurrent' sandpiles must be reachable in this way. Since by construction, our 'identity' sandpile has a value ≥ 3 in each cell before toppling, it will be a 'recurrent' sandpile.
If something is not associative it is not a group. An abelian group is a group which is commutative.
This has causality backwards—being a group requires an identity element. You can't show something is a group without knowing that the identity element exists in the first place.
In fact, a good chunk of how this article talks about the math is just... slightly off.