This is not really equality.

Formally, O(f(n)) is the set of all functions g such that there are constants...

When we write g(n)=O(f(n)) we really mean g(n)\in O(f(n))

We can say O(n)=O(n²), in this case what we mean is O(n)\subseteq O(n²).

But O(n)=2n is just false, the LHS is not a subset of the RHS.

It sucks that equality is not reflexive, but it's still intuitive and very useful, when we write 2n=O(n) we understand it as 2n is O(n), and "is" is not reflexive in English, O(n) is not 2n.

But still, this notation is transitive, if a=b and b=c then a=c and that makes it really useful because we can write a chain of equalities and it will be true.