I can't give you an answer that is well-adapted to Lawson and Michelsohn's formulation of the pseudodifferential calculus. But here's how this sort of argument is supposed to go: two elliptic pseudodifferential operators with the same principal symbol differ by a smoothing operator, and smoothing operators are compact (one often proves this by showing they are Hilbert-Schmidt). By Atkinson's theorem, two Fredholm operators which differ by a compact operator have the same Fredholm index.
So the real point is to understand why the symbol of an operator characterizes it up to smoothing operators. The problem is that you can have a lot of flexibility in exactly what class of operators you decide to allow to be called "pseudodifferential", and so your proof of this fact has to be well-adapted to your specific construction. But I don't think that there is much controversy over the statement that whatever algebra of operators you consider to be the algebra of pseudodifferential operators there should be a symbol map whose kernel consists only of smoothing operators (or, at worst, compact operators). I unfortunately can't help you verify that Lawson and Michelsohn's choice meets this criterion.
EDIT: I see that I did not quite interpret your question correctly. However, showing that the index is determined by the symbol is most of the work toward showing that the index is determined by the symbol class. What I wrote above tells you how to show that there is a well defined index map from the set of symbols to the integers, so as Johannes Ebert points out you only need to check that this map is compatible with the relations which define $K(T^\ast M)$. The relations at the symbol level (homotopy, direct sum) pass to corresponding relations at the level of the operators which do not change the index.