I was looking at the proof of Lagrange's theorem (that the order of a group ##G## is a multiple of the order of any given subgroup ##H##) in Wikipedia:

I understand this proof fine, but I was wondering, instead of finding a bijection between cosets, is it enough to find a bijection between an arbitrary coset ##gH## and the subgroup ##H##? So, for instance, we have a map ##f: gH \rightarrow H##, where ##f(x) = g^{-1}x##. The map is bijective, with inverse ##f^{-1}(y) = gy##. Is there anything wrong with this?

No. Whether you show ##|aH|=|bH|## for arbitrary ##a\, , \, b## or ##|aH|=|H|=|eH|## for all ##a## doesn't make a difference. And ##"="## is transitive. It's simply a matter of taste.