Let $A,S$ be two complete and separable metric spaces, and call $W_1$ the Wasserstein 1-metric and $L$ a positive fixed constant.
Let $p(\cdot|s,a)$ a probability measure on $S$ which satisfies $$\forall s,s'\in S,\ a,a'\in A, \qquad W_1(p(\cdot|s,a),p(\cdot|s',a')) \le Ld_S(s,s')+Ld_A(a,a')$$ And let $\pi(\cdot|s)$ be another probability on $A$ satisfying $$\forall s,s'\in S \qquad W_1(\pi(\cdot|s),\pi(\cdot|s')) \le L d_S(s,s')$$
Can I say that the probability measure on $S$ defined as $$p^\pi(z|s):=\int_A p(z|s,a)\pi(a|s)\ da$$
satisfies $$\forall s,s'\in S \qquad W_1(p^\pi(\cdot|s),p^\pi(\cdot|s')) \le L_\star d_S(s,s')$$
for some other value of $L_\star$ depending on $L$?
I tried with the dual norm but after a ton of computation I didn't get to anything.
I do not have time for a complete answer but might elaborate later if you are still interested. In any case I would rewrite $$p^\pi(z | s) = \int_A p(z| s, a) \lambda(a, a' | s, s')da da'$$ with $\lambda$ being a coupling of the measures $\pi(\cdot, s)$ and $\pi(\cdot, s')$. Then you can use Theorem 4.8 in http://www.cedricvillani.org/sites/dev/files/old_images/2012/08/preprint-1.pdf and optimize over all $\lambda$. This might then lead you to $L_* = 2 L$.