$\newcommand{\Z}{\mathbb Z}\newcommand{\PP}{\mathcal D}\newcommand{\R}{\mathbb R}$Your function $d$ is not a metric, for two reasons: (i) there may be many processes $(X_t)_{t\in\Z}$ with the same distribution $P$ and (ii) your function $d$ does not take into account the values of $X_t$ for negative $t\in\Z$. So, your $d$ is, not a metric, but a pseudometric, which does not allow one to identify limits uniquely.
We can fix these deficiencies as follows: Let $\PP$ denote the set of the distributions of the processes in $\mathcal P$.
Given $P$ and $Q$ in $\PP$, for any natural $m$ let
\begin{equation}
P_m:=P\circ\pi_{-m,\dots,m}^{-1},\quad Q_m:=Q\circ\pi_{-m,\dots,m}^{-1},
\end{equation}
where $\pi_{r,\dots,s}((x_t)_{t\in\Z}):=(x_r,\dots,x_s)$ for any given integers $r,s$ such that $r\le s$.
Let
\begin{equation}
d(P,Q):=\sum_{m=1}^\infty d^{(m)}(P_m,Q_m)2^{-m},
\end{equation}
where $d^{(m)}$ is the Wasserstein metric of order $2$.
We want then to show that $\PP$ is closed with respect to the metric $d$.
Suppose now that we have a sequence $(P^{(n)})$ in $\PP$ such that $d(P^{(n)},Q)\to0$ (as $n\to\infty$) for some probability measure $Q$ (on the cylindrical $\sigma$-algebra) over $\R^\Z$. Then for each natural $m$ we have $d^{(m)}(P^{(n)}_m,Q_m)\to0$. So, which implies thatby the well-known characterization of the convergence in the Wasserstein metric, $P^{(n)}_m\to Q_m$ weakly, $\int_{\R^{\Z_m}} x_t^2\,Q_m(dx)=\lim_n\int_{\R^{\Z_m}} x_t^2\,P^{(n)}_m(dx)<\infty$, and $\int_{\R^{\Z_m}} x_t\,Q_m(dx)=\lim_n\int_{\R^{\Z_m}} x_t\,P^{(n)}_m(dx)=\lim_n0=0$ for $t\in\Z_m:=\{-m,\dots,m\}$, and $P^{(n)}_m\to Q_m$ weakly.
So, $\int_{\R^\Z} x_t\,Q(dx)=0$ and $\int_{\R^\Z} x_t^2\,Q(dx)<\infty$ for all $t\in\Z$, and $P^{(n)}_{r,s}\to Q_{r,s}$ weakly for any given integers $r,s$ such that $r\le s$, where $P^{(n)}_{r,s}:=P^{(n)}\circ\pi_{r,\dots,s}^{-1}$ and $Q_{r,s}:=Q\circ\pi_{r,\dots,s}^{-1}$.
By the stationarity, $P^{(n)}_{r+1,s+1}=P^{(n)}_{r,s}$ for all suitable $r,s,n$. Letting now $n\to\infty$, we conclude that $Q_{r+1,s+1}=Q_{r,s}$, so that $Q$ is the distribution of a stationary process. Also, as we saw, $\int_{\R^\Z} x_t\,Q(dx)=0$ and $\int_{\R^\Z} x_t^2\,Q(dx)<\infty$ for all $t\in\Z$. So, $Q\in\PP$.
We conclude that $\PP$ is closed, as desired.