I do not know P&S textbook, my answer below just reflects some implicit reasoning of some (other) textbooks, not necessarily shared with P&S.
I stress that what I wrote below is not my viewpoint on this very delicate issue.
In a more rigorous perspective, the relevant object is $\hat{\phi}(f)$ that is formally interpreted as
$$\hat{\phi}(f):= \int \hat{\phi}(x) f(x) d^4x$$
It is the field operator *smeared by the test function* $f$, any smooth compactly supported real function $f$ defined in the spacetime.
In fact, $\hat{\phi}(f)$ is a densely defined Hermitian operator in the Hilbert space for every such $f$ (*).
Technically speaking, the associated observable is the closure of that operator which is selfadjoint and thus, in principle, is an observable. However there is a dense invariant domain in common for all operators $\hat{\phi}(f)$ when varying $f$, and the operator closure on that domain defines the said selfadjoint operator. I will assume to deal with this domain henceforth.
The commutation relations written into a rigorous version are $$[\hat{\phi}(f), \hat{\phi}(g)]= 0\quad \mbox{if the supports of f and g are causally separated}$$
Here we are considering the commutation relation of **proper observables**. These represent observables localised in regions defined by the supports of the two smearing functions. In turn, these sets are causally separated.
These observables are expected to be compatible since, for "obvious causal reasons" *at least in some folklore of QFT* (see below), their measurements cannot "disturb each other".
Compatibility, in the standard version of QM, is equivalent to commutativity of the spectral measures which, on suitable invariant domains implies commutativity of the same observables. This justifies the above commutation relations.
Formally, we can also write
$$0= [\hat{\phi}(f), \hat{\phi}(g)] = \int\int [\hat{\phi}(x), \hat{\phi}(y)] f(x)g(y) d^4xd^4y.$$
Arbitrariness of $f$ and $g$ should imply
$$[\hat{\phi}(x), \hat{\phi}(y)]=0$$
when $x$ and $y$ are causally separated, though this identity is just a shortened version of its rigorous version written above.
**ADDENDUM**. I sketch here a proof based on some folklore assumptions. I explicitly admit that they are disputable, especially the second and the third one, on account of the modern theory of quantum measurement. I do not know if there is a better way to justify the commutativity postulate according to the modern view.
(On my side, I just assume commutativity since it produces interesting and important physical facts.)
Let us assume that:
1) real smeared boson fields (selfadjoint operators) are observables;
2) measurements of them are described in terms of the *Luders projection postulate* using the spectral measures of the smeared fields;
3) outcomes of the above measurements are recorded in spacetime regions inside the support of the smearing functions.
Consider two causally separated regions in Minkowski spacetime $\Omega, \Omega' \subset M^4$ and two orthogonal projectors $P^{(\Omega)}_E$, $P^{(\Omega')}_{E'}$ of the spectral measures of $\hat{\phi}(f)$ and $\hat{\phi}(f')$ respectively, where $supp(f) \subset \Omega$ and $supp(f')\subset \Omega'$.
Consider a pure state of the quantum field $\Psi$. After the measurement of $\hat{\phi}(f)$, according to (2), the post measurement state is $P^{(\Omega)}_E\Psi$ (up to normalization) if the outcome was $E$.
Therefore, the probability to next obtain $E'$ when measuring $\hat{\phi}(f')$ is $||P_{E'}^{(\Omega')}P^{(\Omega)}_E\Psi||^2$.
**On the other hand, the temporal (causal) order of the two measurements (more properly of the recording procedures of the outcomes) cannot make sense, since they are recorded in spatially separated regions.**
**Therefore, if swapping the measurements, one should obtain the same result in terms of probability (observed frequency).**
The conclusion is that
$$||P^{(\Omega)}_EP_{E'}^{(\Omega')}\Psi||^2= ||P_{E'}^{(\Omega')}P^{(\Omega)}_E\Psi||^2$$
In other words, if $\rho_\Psi:= |\Psi\rangle \langle \Psi|$,
$$tr\left(\rho_\Psi P^{(\Omega)}_EP_{E'}^{(\Omega')}\right) = tr\left(\rho_\Psi P_{E'}^{(\Omega')}P^{(\Omega)}_E\right)\:.$$
The result extends by linearity and continuity to every trace class operator $\rho$ in place of $\rho_\Psi$. As a consequence
$$P_{E'}^{(\Omega')}P^{(\Omega)}_E = P^{(\Omega)}_E P_{E'}^{(\Omega')}$$
The spectral measures of $\hat{\phi}(f)$ and $\hat{\phi}(f')$ therefore **commute**.
As a consequence, paying attention to domains,
$$[\hat{\phi}(f), \hat{\phi}(f')]=0\:.$$
As an alternative approach, but leading to the same result, suppose that the measurement in $\Omega$ **is not selective**. So we test $P_E^{(\Omega)}$ and $\neg P_E^{(\Omega)}:= I- P_E^{(\Omega)}$ without knowing the result. If the generically mixed initial state is $\rho$ the post measurement state is
$$\rho' := P_E^{(\Omega)} \rho P_E^{(\Omega)} + (I-P_E^{(\Omega)})\rho (I-P_E^{(\Omega)})\:.$$
The probability to measure $E'$ is therefore
$$tr\left(P^{(\Omega')}_{E'} \rho'\right)\:.$$
**However, since this measurement is located in a causally separated region, the same probability should arise when performing the measurement on the initial state $\rho$: That is because there is an observer who describes the measurement in $\Omega'$ **before** the one in $\Omega$**.
Hence we are committed to assume that
$$tr\left(P^{(\Omega')}_{E'} \rho'\right)= tr\left(P^{(\Omega')}_{E'}) \rho'\right)$$
An easy computation based on linearity and the cyclic property of the trace yields
$$tr\left(\rho \left(P^{(\Omega')}_{E'} - P_{E}^{(\Omega)} P^{(\Omega')}_{E'} P_{E}^{(\Omega)} - (I-P_{E}^{(\Omega)}) P^{(\Omega')}_{E'} (I-P_{E}^{(\Omega)})\right)\right)=0\:.$$
Arbitrariness of $\rho$ entails
$$P^{(\Omega')}_{E'} = P_{E}^{(\Omega)} P^{(\Omega')}_{E'} P_{E}^{(\Omega)} + (I-P_{E}^{(\Omega)}) P^{(\Omega')}_{E'} (I-P_{E}^{(\Omega)})\:.$$
Applying $P_E^{(\Omega)}$ separately on both sides produces:
$$ P_E^{(\Omega)}P^{(\Omega')}_{E'} = P_E^{(\Omega)}P_{E}^{(\Omega)} P^{(\Omega')}_{E'} P_{E}^{(\Omega)} +0 = P_{E}^{(\Omega)} P^{(\Omega')}_{E'} P_{E}^{(\Omega)} \:.$$
and
$$P^{(\Omega')}_{E'} P_E^{(\Omega)} = P_{E}^{(\Omega)} P^{(\Omega')}_{E'} P_{E}^{(\Omega)}P_E^{(\Omega)} +0 = P_{E}^{(\Omega)} P^{(\Omega')}_{E'} P_{E}^{(\Omega)} \:.$$
So that we have again the thesis:
$$P^{(\Omega')}_{E'} P_E^{(\Omega)}=P_E^{(\Omega)}P^{(\Omega')}_{E'} .$$
If we do **not** assume that $\hat{\phi}(f)$ is directly measurable, but other *formal local observables* generated by it, e.g. the *smeared renormalized stress energy tensor*, are, we can repeat the proof above for them.
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(*) I stress that the smeared fields $\hat{\phi}(f)$, for **real boson quantum fields** are considered **observables** in a the perspective of Local Quantum Physics, the view relying on the *Haag-Kastler formulation* of QFT. Other fields, like fermions, are not considered observables.
Abstractly speaking, They are elements of the $*$-algebra of observables of the theory (their exponentials $e^{i\hat{\phi}(f)}$ define the generators of the Weyl $C^*$-algebra of quasi local observables).