Since the path-integral formulation is manifestly Lorentz symmetric, we only need to show that $\phi(\vec x,t)$ and $\phi(\vec y,t)$ commute when $\vec x\neq \vec y$. If they commute when the times are equal, then the path integral's manifest Lorentz symmetry implies that they commute for all spacelike separations.
In the canonical formulation with operators on a Hilbert space, two different kinds of ordering are involved: operator order, and time order. Nothing prevents us from considering quantities like $\phi(\vec x,t)\phi(\vec x',t')|0\rangle$ with $t'>t$, where the time order is different than the operator order.
In contrast, the path integral always gives us time-ordered correlation functions. It doesn't come with any separate notion of operator order, but we can still obtain equal-time commutation relations using the following identity. Let $X$ and $Y$ be arbitrary products of field operators whose time-arguments are all $>t$ and $<t$, respectively, with enough margin to avoid interfering with the $\epsilon\to 0$ limit shown below, where $\epsilon$ is positive. Then
\la 0|X[\phi(\vec x,t),\phi(\vec y,t)]Y|0\ra
\la 0|X\phi(\vec x,t+\epsilon)\phi(\vec y,t)Y|0\ra
\la 0|X\phi(\vec y,t)\phi(\vec x,t-\epsilon)Y|0\ra
\la 0|T\phi(\vec x,t+\epsilon)\phi(\vec y,t)XY|0\ra
\la 0|T\phi(\vec x,t-\epsilon)\phi(\vec y,t)XY|0\ra
The operators $X$ and $Y$ are arbitrary (as long as they're not too close to the time $t$), so this gives us the commutator $[\phi(\vec x,t),\phi(\vec y,t)]$ sandwiched between arbitrary states, which determines the commutator itself as an operator relation.
When $\vec x\neq\vec y$, the two terms on the last line are clearly equal to each other in the limit $\epsilon\to 0$. That implies $[\phi(\vec x,t),\phi(\vec y,t)]=0$ for $\vec x\neq \vec y$, and the path integral's manifest Lorentz symmetry then implies that they commute for all spacelike separations.
A similar approach can be used to derive the equal-time commutation relation $[\phi(\vec x,t),\pi(\vec y,t)]\propto \delta(\vec x-\vec y)$, where $\pi$ is the canonical conjugate of $\phi$. This takes more work, but it isn't necessary for answering the question, because (if $\pi=\dot\phi$) it follows easily from the unequal-time commutator of $\phi(x)$ and $\phi(y)$ when $\vec x\neq \vec y$, which we already deduced.