$\newcommand\R{\mathbb R}\newcommand\B{\mathcal B}\newcommand\Si{\Sigma}\newcommand\ga{\gamma}$
Your conjecture is false in general.
E.g., suppose that the underlying measurable space
$(S,\Si)$ on which $P$ and $Q$ are defined is $(\R,\B(\R))$, where $\B(\R)$ is the Borel $\sigma$-algebra over $\R$. Let $P$ be the uniform distribution on $(0,1)$. Then any distribution $R$ on $(\R,\B(\R))$ is of the form $T_{\#}P$ for some $T$; namely, $T$ is the quantile transformation given by the formula
$$T(u)=H^{-1}(u):=\inf\{x\in\R\colon H(x)\ge u\}$$
for $u\in(0,1)$, where $H$ is the cdf of the distribution $R$. Let now $Q$ be any distribution on $(\R,\B(\R))$ other than $P$. There are real-valued random variables (r.v.'s) $X$ and $Y$ with respective distributions $P$ and $Q$ such that $W^2(P,Q)=E(X-Y)^2$; for instance, take any r.v. $X\sim P$ and let then $Y:=G^{-1}(X)$, where $G$ is the cdf of $Q$. Let now $Z:=(X+Y)/2$ and let $R$ be the distribution of $Z$. Then the right-hand side of your conjectured identity is
$$\le d^2(P,R) + W^2(R,Q)\le E(X-Z)^2+E(Z-Y)^2=\frac12\,E(X-Y)^2=\frac12\,W^2(P,Q)<W^2(P,Q);$$
the latter inequality holding because $Q\ne P$. So, the right-hand side of your conjectured identity is strictly less than its left-hand side.
A correct version of your conjectured identity is a direct application of the triangle inequality in $L^2$, as follows:
$$W(P,Q)=\inf_T[d(P,T_{\#}P)+W(T_{\#}P,Q)].\tag{1}$$
Indeed,
$$d(P,T_{\#}P)+W(T_{\#}P,Q)=\inf_\ga\Big[\sqrt{\int_{S\times S}(x-T(x))^2\ga(dx,dy)}+\sqrt{\int_{S\times S}(T(x)-y)^2\ga(dx,dy)}\Big]
\ge\inf_\ga\sqrt{\int_{S\times S}(x-y)^2\ga(dx,dy)}=W(P,Q),$$
where $\inf\limits_\ga$ is taken over all couplings $\ga$ of $P$ and $Q$. So, the right-hand side of (1) is no less than its left-hand side. On the other hand, taking $T$ to be the identity map of $S$, we see that the right-hand side of (1) is no greater than its left-hand side. Thus, (1) is proved.
