Extension of spectral gap inequality in Wasserstein distance Let $E$ be a separable $\mathbb R$-Banach space, $\rho_r$ be a metric on $E$ for $r\in(0,1]$ with $\rho_r\le\rho_s$ for all $0<r\le s\le1$, $\rho:=\rho_1$, $$d_{r,\:\delta,\:\beta}:=1\wedge\frac{\rho_r}\delta+\beta\rho\;\;\;\text{for }(r,\delta,\beta)\in[0,1]\times(0,\infty)\times[0,\infty)$$ and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$.

Assume we arbe able to show that for all $n\in\mathbb N$ there is a $\alpha\in[0,1)$ and $(r,\delta,\beta)\in[0,1]\times(0,\infty)\times(0,1)$ with$^1$ $$\operatorname W_{d_{r,\:\delta,\:\beta}}\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha\operatorname W_{d_{r,\:\delta,\:\beta}}\left(\delta_x,\delta_y\right)\tag1$$ for all $x,y\in E$, where $\delta_x$ denotes the Dirac measure on $(E,\mathcal B(E))$ at $x\in E$. Why are we able to conclude that there is a $(c,\lambda\in[0,\infty)^2$ with $$\operatorname W_\rho\left(\nu_1\kappa_t,\nu_2\kappa_t\right)\le ce^{-\lambda t}\operatorname W_\rho\left(\nu_1,\nu_2\right)\tag2$$ for all $\nu_1,\nu_2\in\mathcal M_1(E)$ and $t\ge0$?

It's clear to me that if $\kappa$ is any Markov kernel on $(E,\mathcal B(E))$ and $d$ is any metric on $E$ such that there is a $\alpha\ge0$ with $\operatorname W_d\left(\delta_x\kappa,\delta_y\kappa\right)\le\alpha\operatorname W_d\left(\delta_x,\delta_y\right)$ for all $x,y\in E$, then this extends to $\operatorname W_d(\mu\kappa,\nu\kappa)\le\alpha\operatorname W_d(\mu,\nu)$ for all $\mu,\nu\in\mathcal M_1(E)$. Moreover, it's clear that $\operatorname W_d\left(\delta_x,\delta_y\right)=d(x,y)$.
Note that for any choice of $(r,\delta,\beta)\in[0,1]\times(0,\infty)\times[0,\infty)$, it holds $$\beta\rho\le d_{r,\:\delta,\:\beta}\le\left(\frac1\delta+\beta\right)\rho.\tag3$$
Remark: The desired claim seems to be used in the proof of Theorem 3.4 in https://arxiv.org/pdf/math/0602479.pdf.

$^1$ If $(E,d)$ is a complete separable metric space and $\mathcal M_1(E)$ is the space of probability measures on $\mathcal B(E)$, then the Wasserstein metric $\operatorname W_d$ on $\mathcal M_1(E)$ satisfies the identity $$\operatorname W_d(\mu,\nu)=\sup_{\substack{f\::\:E\:\to\:\mathbb R\\|f|_{\operatorname{Lip}(d)}\:\le\:1}}(\mu-\nu)f\;\;\;\text{or all }\mu,\nu\in\mathcal M_1(E),$$ where $$|f|_{\operatorname{Lip}(d)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{d(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$ and $\mu f:=\int f\:{\rm d}\mu$ for $\mu$-integrable $f:E\to\mathbb R$.
 A: I can answer assuming some regularity on the Markov semigroup, which I would expect to be satisfied in most cases. Specifically, assume local (in time) Lipschitz continuity on your Markov semigroup, i.e. 
$$\forall s_0>0, \exists C>0, \forall s\in[0,s_0], \forall \mu_1,\mu_2 : \mathrm{W}(\mu_1\kappa_s,\mu_2\kappa_s)\le C\mathrm{W}(\mu_1,\mu_2)$$ 
(I do not precise for which metric, since the two metrics under consideration are Lipschitz-equivalent and so only the constant $C$ would change when passing from one to the other.)
Using convexity of Wasserstein distance, every Lipschitz/contraction bound we have on Dirac masses is also true for arbitrary measures (I guess that is what you mean at the end of your question, although an $\alpha$ appears to be missing).
For any $t_0$, using (1)  with $n=1$ iteratively and the double inequality (3):
\begin{align*}
\mathrm{W}_\rho(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0}) 
  &\le \frac1\beta \mathrm{W}_{d_{r,\delta,\beta}}(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0}) \\
  &\le \frac{\alpha^{t_0}}{\beta} \mathrm{W}_{d_{r,\delta,\beta}}(\delta_x,\delta_y) \\
  &\le \alpha^{t_0}\Big(\frac{1}{\beta\delta}+1\Big) \mathrm{W}_\rho(\delta_x,\delta_y)
\end{align*}
Since $\alpha\in(0,1)$, this is what you needed.
(Side note: this kind of computation shows that any decay of the form
$$ d(T^n(x),T^n(y)) \le f(n) d(x,y)$$
where $d$ is any metric, $T$ is any Lipschitz dynamical system, and $f(n) \to 0$ as $n\to \infty$ (or even $f(n)<1$ for some $n$), actually imply exponential decay. This is pretty basic, but seems to be sometimes overlooked.)
A: Building up on Benoît Kloeckner's answer, consider the following simplified scenerio: Let $(E,d)$ be a complete separable metric space, $(\kappa_t)_{\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$ with $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,1)\tag{10}$$ for some $c\ge0$ and $$\operatorname W_d(\delta_x\kappa_1,\delta_y\kappa_1)\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag{11}$$ for some $\alpha\in(0,1)$.
From $(11)$, we easily deduce $$\operatorname W_d\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag{12}$$ for all $x,y\in\mathbb N$ and $n\in\mathbb N_0$. If $t>0$, we may write $t=n+r$ for some $n\in\mathbb N_0$ and $r\in[0,1)$ so that $$\operatorname W_d\left(\delta_x\kappa_t,\delta_y\kappa_t\right)\le\alpha^n\operatorname W_d\left(\delta_x\kappa_r,\delta_y\kappa_r\right)\le c\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag{13}$$ for all $x,y\in E$ by $(12)$ and $(10)$.
Now we only need to note that $$c\alpha^n=\frac c\alpha\alpha^{n+1}\le\frac c\alpha\alpha^t\tag{14}$$ (the last "$\le$" is actually a "$<$" as long as $c\ne0$) and hence we obtain $$\operatorname W_d\left(\mu\kappa_t,\nu\kappa_t\right)\le\tilde ce^{-\lambda t}\operatorname W_d(\mu,\nu)\tag{15}$$ for all $\mu,\nu\in\mathcal M_1(E)$, where $$\tilde c:=\frac c\alpha$$ and $$\lambda:=-\ln\alpha.$$
Remark
I'd still be interested in the question whether this result still holds when $(10)$ and $(11)$ are replaced by the following assumption: There is a $t_0>0$ with $$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,t_0)\tag{10'}$$ and $$\operatorname W_d(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0})\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag{11'}$$ for some $\alpha\ge0$.
(The original statement in this answer is the particular case $t_0=1$.)
