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 $(E,\mathcal B(E))$.



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> 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 $t\ge0$?
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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>.




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$^1$ 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 $f:E\to\mathbb R$.



Extension of spectral gap inequality in Wasserstein distance
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.)


