Right. If $L = U$ is the lattice of the quadratic form $u(x,y) = 2 xy,$ and $M,N$ are positive definite, the conclusion is that $M,N$ are in the same genus. That is, they are rationally equivalent "without essential denominator." There is no complete proof printed in one place. I first saw this on page 378 of SPLAG by Conway and Sloane, first edition. The observation may be due to Conway. This is a small part of finding certain automorphism groups, and is first apparent in the articles on the automorphism group of the Leech Lattice. Anyway, click on my name and just go through my question with promising titles. In a minute I will find the one with a sketch of a proof, put a link here.
Found it, http://mathoverflow.net/questions/70666/lorentzian-characterization-of-genus
I also checked with Wai Kiu Chan about the case of "odd" lattices such as the sum of squares, it turns out it does not matter, same outcome.
Meanwhile, it is exactly this observation that allows one to conclude, given a positive "even" lattice with covering radius strictly below $\sqrt 2,$ such as $\mathbb E_8,$ that there is only one class in the genus, i.e. that your integral cancellation holds. See http://mathoverflow.net/questions/69444/a-priori-proof-that-covering-radius-strictly-less-than-sqrt-2-implies-class-nu