Let me first note that $[H,N]=0$ is a purely mathematical relation, which can be verified for the Hamiltonian that you are working with. Derivations in (quantum) statistical mechanics textbooks often assume that it is the case, while leaving the exceptions from the rule to the chapters about superconductivity and superfluidity. Another purely mathematical fact is that $\mu$ is a number, which commutes with either $H$ or $N$.
The quantum mechanical meaning of equality $[H,N]=0$ is that the energy and the particle number can be used together as quantum numbers to label the states. (There are obviously more quantum numbers, but these two are privileged in the context of the statistical mechanics.) Switching from the canonical to the grand canonical ensemble doesn't change anything in the Hamiltonian or its states, but changes the states that we include in the statistical sum: in the canonical ensemble they all have the same $N$, whereas in the grand canonical ensemble different values of $N$ are allowed.
In other words non-conservation of particles has different meaning in QM and Stat. Mech.: in the former it means that the energy states cannot be characterized by a particle number, in the latter - that we consider states with different numbers of particles.
Finally, note that this is not a quantum mechanical feature: although there is no semantic confusion in the classical case, the classical Hamiltonians do conserve the number of particles, as they have a fixed number of degrees of freedom. Thus, when we work in the grand canonical ensemble, we use a different Hamiltonian for every $N$, e.g.,