Why aren't four-vectors used in the definition of a Klein-Gordon quantum field?
As I mentioned in the comments, P&S are working in the Schrodinger picture which means that the operator fields are time-independent. Of course, in the Heisenberg picture, the solution of the Klein-Gordon equation is dependent on time (and then it will have four-vectors). In order to see this, let us write down the Klein-Gordon equation: \begin{equation} \left(\partial^2 + m^2 \right) \phi(x)=0 \end{equation} where $g=\mathrm{diag}(+1,-1,-1,-1)$. Then the solutions in the Heisenberg picture can be written as: \begin{equation} \phi(x) = e^{\pm i p_\mu x^\mu} \end{equation} which can be easily verified: \begin{equation} \begin{aligned} \partial^2 \phi & = \partial_\mu \partial^\mu \left(e^{\pm i p_\nu x^\nu}\right) \\& = \partial_\mu \left(\pm i p^\mu\right) \left(e^{\pm i p_\nu x^\nu}\right) \\& = \left(\pm i p^\mu\right) \left(\pm i p_\mu\right) e^{\pm i p_\nu x^\nu} \\& = - p_\mu p^\mu e^{\pm i p_\nu x^\nu} \\& = -(E^2 - \mathbf{p}^2) e^{\pm i p_\nu x^\nu} \\& = -m^2 e^{\pm i p_\nu x^\nu} \\& = -m^2 \phi \end{aligned} \end{equation} and so: \begin{equation} \left(\partial^2 +m^2\right)\phi = \left(-m^2 +m^2\right)\phi = 0 \end{equation} It is normal to write the solution in terms of positive frequency solutions and negative frequency solutions: \begin{equation} \phi(x)=\phi_+(x) + \phi_-(x) = a e^{- i p_\nu x^\nu} + b e^{+ i p_\nu x^\nu} \tag{1} \end{equation} Of course, we also need to sum over all energy-momentum values $p_\mu$ (because equation $(1)$ is a solution for any value of $p_\mu$). Hence, the general solution is: \begin{equation} \phi(\mathbf{x},t) = \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + b(\mathbf{p}) e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}}\right] \end{equation} where $N$ is a normalization constant.
In order to see how to switch between the Dirac and Schrodinger picture, I refer you to section $2.4$ of P&S.
Edit I couldn't help my self and will quickly add this:
P&S are discussing the real Klein-Gordon field, which means: $$ \phi = \phi^* $$ and so: \begin{equation} \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i p^\mu x_\mu} + b(\mathbf{p}) e^{ip^\mu x_\mu}\right] = \int \frac{\mathrm{d}^3 \mathbf{p}}{N^*} \; \left[ a^*(\mathbf{p}) e^{ ip^\mu x_\mu} + b^*(\mathbf{p}) e^{-i p^\mu x_\mu}\right] \end{equation} which implies: \begin{equation} \begin{array}{cc} a(\mathbf{p}) = b^*(\mathbf{p}) \; ,& b(\mathbf{p}) = a^*(\mathbf{p}) \end{array} \end{equation} and $N$ must be a real. So the real field can be written as: \begin{equation} \phi(\mathbf{x},t) = \int \frac{\mathrm{d}^3 \mathbf{p}}{N} \; \left[ a(\mathbf{p}) e^{- i E_{\mathbf{p}} t + i \mathbf{p} \cdot \mathbf{x}} + a^*(\mathbf{p}) e^{i E_{\mathbf{p}} t - i \mathbf{p} \cdot \mathbf{x}}\right] \end{equation}