I saw this question: What does "formal" mean? But I don't think it answers my question.
In Hatcher's book "Vector Bundles and K-Theory", he defines $K^0(X)$ to be the group of formal differences of vector bundles over the space $X$, with the equivalence relation that $E_1-E_1'=E_2-E_2'$ iff $E_1\oplus E_2\cong_s E_2\oplus E_1'$, where $\cong_s$ means they are stably isomorphic. By defining it like this, it can easily be turned into an abelian group.
Are these differences meaningless and form part of a trick to get the group to work, or do they actually hold some sort of topological significance?
Well, I'm not sure what you mean by "meaningless", but these "differences" are just symbols. You could consider $E_1-E_1'$ as just a notation for the ordered pair $(E_1,E_1')$. The group $K^0(X)$ is then defined as the set of equivalence classes of ordered pairs of vector bundles on $X$ under the equivalence relation $(E_1,E_1')\sim(E_2,E_2')$ iff $E_1\oplus E_2'\cong_s E_2\oplus E_1'$. It then makes sense to write the equivalence class of an ordered pair $(E_1,E_1')$ as $E_1-E_1'$ because this operation behaves formally like subtraction. (And in fact, it is the universal way to turn the monoid of vector bundles up to isomorphism under $\oplus$ into a group.)
It is perhaps instructive to observe that if you define an equivalence relation on $\mathbb{N}^2$ by $(a,b)\sim (c,d)$ iff $a+d=c+b$, then the equivalence classes can naturally be identified with the integers, sending $(a,b)$ to the integer $a-b$. Here we're doing the same sort of thing to abstractly allow ourselves to subtract vector bundles, by representing the result of such a subtraction as an equivalence class of pairs of vector bundles.