Consider the simple harmonic oscillator Hamiltonian
$$\tag{1} \hat{H}~:=~\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^2 ~=~\hbar\omega(\hat{n}+\frac{1}{2}),$$
where $\hat{n}:=\hat{a}^{\dagger}\hat{a}$ is the number operator.
Let us put the constants $m=\hbar=\omega=1$ to one for simplicity. Then the annihilation and creation operators are
$$\tag{2} \hat{a}~=~\frac{1}{\sqrt{2}}(\hat{x} + i \hat{p}), \qquad \hat{a}^{\dagger}~=~\frac{1}{\sqrt{2}}(\hat{x} - i \hat{p}),$$
or conversely,
$$\tag{3} \hat{x}~=~\frac{1}{\sqrt{2}}(\hat{a}^{\dagger}+\hat{a}), \qquad \hat{p}~=~\frac{i}{\sqrt{2}}(\hat{a}^{\dagger}-\hat{a}).$$
Eqs. (2) and (3) yield the identity
$$\tag{4} [\hat{x},\hat{p}]~=~i[\hat{a},\hat{a}^{\dagger}]. $$
In other words, the non-commutativity of the ladder operators $\hat{a}$ and $\hat{a}^{\dagger}$ is directly related to the non-commutativity in the canonical commutation relation (CCR):
$$\tag{5} [\hat{x},\hat{p}]~=~i{\bf 1}\qquad \Longleftrightarrow \qquad [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}. $$
II) Thus if the ladder operators $\hat{a}$ and $\hat{a}^{\dagger}$ would commute, as OP ponders, then all operators of the theory would commute, all quantum mechanics would have been thrown out with the bath water, and we would be back doing classical mechanics.
Phrased equivalently, if the ladder operators $\hat{a}$ and $\hat{a}^{\dagger}$ would commute, then $\hat{a}$ (and $\hat{a}^{\dagger}$) would be a normal operator, and it would indeed be possible to associate $\hat{a}$ with a complex observables, cf. thisthis Phys.SE answer. The corresponding two commuting self-adjoint operators would in this case be $\hat{x}$ and $\hat{p}$, which then could be measured simultaneously.
In conclusion: As we know that the ladder operators $\hat{a}$ and $\hat{a}^{\dagger}$ do not commute, then $\hat{a}$ is not a normal operator, and hence $\hat{a}$ is not a complex observable.
