No, the first set has an element $(2,1)$ which is not the smallest element of the set but has no immediate predecessor, while the second set does not have such an element (therefore, there is no order preserving bijection, as if there is, say $f$ , then $f(2,1)$ cannot be the smallest element, hence, it must have a predecessor $p$ , but then $f^{-1}(p)<(2,1)$ , that is $f^{-1}(p)=(1,x)$ , and $p<f(1,x+1)<f(2,1)$ ).