Is the commutation of all possible operators sufficient to identify a spacelike interval? It has been claimed (e.g. here) and apparently already been established, that the interval $x - y$ being (called) "spacelike" implies that $\bigl[\hat O (x),\, \hat O' (y)\bigr]=0$ for any two (not necessarily distinct) operators $\hat O$ and $\hat O'$ corresponding to physical observables evaluated at $x$ or at $y$, respectively: 
$$\text{spacelike}( \, x - y \, ) \quad\implies\quad \Bigl( \forall \hat O \, \forall \hat O': \Bigl[\hat O (x),\, \hat O' (y)\Bigr] = 0 \Bigr). $$
Is the converse correct, too, that the vanishing commutators imply (or are sufficient for) the interval $x - y$ to be (called) "spacelike":
$$\Bigl(\forall \hat O \, \forall \hat O':  \Big[\hat O (x),\, \hat O' (y)\Bigr] = 0 \Bigr) \quad\implies\quad \text{spacelike}( \, x - y \, ) \, ?$$
 A: As far as I know, all quantum theories have non-compatible observables (i.e., their corresponding commutators do not vanish as long as the space-time points at which they are evaluated are causally related), therefore if at two given points every pair of observables commute, then these two point are separated by a space-like interval. This should actually be the intrinsic way in which one defines the light-cone in a quantum field theory (quantum theories which do not have local observables are perhaps subtle in this issue and I ignore how —if possible— one can define the light-cone in an intrinsic manner in these theories).
So yes, it seems the implication works in both ways.
A: No, any operators that can be measured simultaniously in quantum mechanics will commute, eg two different components of an electric field, these commute at non-spacelike separation, excluding electron-positron effects.
A: The meaning of the causal commutation rules $[O_1(x),O_2(y)] = 0$ if $x$ and $y$ are spacelike separated is that pairs of fields can be prepared independently at the same time, with time interpreted in the reference frame of an arbitrary observer.
The assertion in your converse question is not true in conformal theories in even dimension, where as a form of Huyghen's principle, (necessarily massless) fields commute at spacelike and timelike separation. 
On the other hand, Poincare covariance implies that even for a single massive field, $[O(x),O(y)]$ is zero in a full neighborhood of two points only if the latter are spacelike separated.
(I had claimed in the first version of my answer that a scalar field and its gradient don't commute at spacelike separation, but this is false, as can be seen by differentiation of the commutation relation with respect to $x$ or $y$.
A more realistic example is QED, where the vector potential $A(x)$ is unobservable, whereas its exterior derivatives define the observable electric field $E(x)$ and the observable magnetic field $B(x)$, and these can indeed be prepared independently at the same time.)
