Yes well, shortly after I posted #3 I realized that the redefined addition doesn't satisfy commutativity (and also probably not associativity) but I left my post as it was because as I said at start I believe there are infinite (if not infinite just too many ) ways to redefine the operations, I just presented one (failed) attempt to redefine addition.
what about if we redefine addition as multiplication, I mean $$(x_1,y_1)\oplus (x_2,y_2)=(x_1x_2,y_1y_2)$$ then this I think satisfies commutativity and associativity , the neutral element becomes (1,1) (instead of 0,0) and the opposite of (x,y) is (1/x,1/y). Is then ##R^2-(0,0)## a vector space under this redefined addition (and the usually defined multiplication by scalar)? What do you think
@pasmith ?