Let $E$ be a separable $\mathbb R$-Banach space, $\rho$ be a complete separable metric on $E$, $\operatorname W_\rho$ denote the Wasserstein metric of order $1$ associated to $\rho$, $\mathcal M_1(E)$ denote the set of probability measures on $(E,\mathcal B(E))$ and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ with $$\operatorname W_\rho(\mu\kappa_t,\nu\kappa_t)\le ce^{-\lambda t}\operatorname W_\rho(\mu,\nu)\;\;\;\text{for all }\mu,\nu\in\mathcal M_1(\mu,\nu)\tag1$$ for some $c\ge0$ and $\lambda>0$.

Are we able to conclude that $(\kappa_t)_{t\ge0}$ has a unique invariant measure $\mu_\ast\in\mathcal M_1(E)$?

By $(1)$, the adjoint semigroup $(\kappa_t^\ast)_{t\ge0}$ is eventually contractive: Let $t_0\ge0$ with $$ce^{-\lambda t}<1\;\;\;\text{for all }t\ge t_0$$ and $t\ge t_0$. Since the Wasserstein space $$\mathcal S^1(E,\rho):=\left\{\mu\in\mathcal M_1(E):(\mu\otimes\delta_0)\rho<\infty\right\}$$ equipped with $\operatorname W_\rho$ is complete and hence we can apply Banach's fixed-point theorem yielding that there is a unique $\mu_\ast\in\mathcal S^1(E,\rho)$ with $$\mu_\ast\kappa_t=\mu_\ast\tag2.$$ Moreover, for any $\mu_0\in\mathcal S^1(E,\rho)$ and $$\mu_n:=\mu_{n-1}\kappa_t\;\;\;\text{for }n\in\mathbb N,$$ it holds $$\operatorname W_\rho(\mu_n,\mu_\ast)\xrightarrow{n\to\infty}0\tag3.$$

So, all what's left to prove is that $\mu_\ast$ does not depend on $t$, i.e. $\mu_\ast$ is invariant with respect to $\kappa_t$ for all $t\ge t_0$.

BTW: Is this all we can hope for or can we even conclude that $\mu_ast$ must be invariant with respect to $\kappa_t$ for all $t\ge\color{red}0$?