The classical poisson bracket with the generator of any transformation gives the infinitesimal evolution with respect to that transformation. The familiar
$$ \partial_t f = \{H,f\}$$
means nothing else than the time evolution of any observable is given by its Poisson bracket with the Hamiltonian, which is the generator of time translation. More generally, given any element $g$ of the Lie algebra of observables on the phase space, the infinitesimal evolution under the transformation that it generates (parametrized by an abstract "angle" $\phi$) is given by
$$ \partial_\phi f = \{g,f\}$$
What is meant by this is that every observable $f$ gives rise (by Lie integration) to a symplectomorphism $\mathrm{exp}(\phi f)$ on the phase space (since the true Lie integration of observables projects down surjectively onto the symplectomorphisms). More precisely, the statement above therefore reads
$$ \partial_\phi( f \circ \mathrm{exp}(\phi g))\rvert_{\phi = 0} = \{g,f\}$$
so that vanishing Poisson bracket implies $f \circ \mathrm{exp}(\phi g) = f$, i.e. invariance of the observable under the induced symplectomorphism.
Thus, if the Poisson bracket of $f$ and $g$ vanishes, that means that they describe an infinitesimal transformation that is a symmetry for the other.
For example, an observable is invariant under rotation if its Poisson bracket with all $L^i = \epsilon^{ijk}x^jp^k$ (components of $\vec L = \vec x \times \vec p$) vanishes.
For $x$ and $p$, the non-vanishing $\{x,p\}$ means the rather trivial insight that $x$ is not invariant under the translations generated by the momentum.
Fun fact: Wondering what the actual Lie integration of these infinitesimal symmetries might be leads directly to the quantomorphism group, and is a natural starting point for geometric quantization, as discussed in this answer.
Responding to the tangential question in a comment:
Considering the invariant under generation idea, so when trying to apply Noether's theorem, one only has to apply the Poisson bracket of the Lagrangian (e.g.) with $x$, and if it vanishes momentum is conserved for that system?
No, it means that the "Lagrangian" (you cannot really take the Lagrangian, because it is not a function on the phase space) is invariant under varying the momentum of the system (since $x$ generates "translations" in momentum). The Hamiltonian equivalent of Noether's theorem is simply that $f$ is conserved if and only if
$$ \{H,f\} = 0$$
since conservation means invariance under time evolution.
