I try to prove that $P_t := e^{\lambda t (P-I)}$ (where $Pf:= \int f(y) P(\cdot , dy)\in \mathcal{C}_0(\mathbb{R}^d)$, for $f\in \mathcal{C}_0(\mathbb{R}^d)$), is a strongly-continuous contraction semigroup (on $(\mathcal{C}_0(\mathbb{R}^d),||\, .\, ||_{\infty})$).
But the condition $P_{t+s}=P_tP_s$ seems to me to be understood as $$P_{t+s}=P_t\circ P_s,$$ so you have to prove, for $f \in \mathcal{C}_0(\mathbb{R}^d)$ that $P_{t+s}(f)=P_t(P_s(f))$ ; here I have : $$\begin{eqnarray} P_{t+s}f &=&e^{\lambda (t+s) (Pf-f)} \\ &= &e^{\lambda t(Pf-f) + \lambda s(Pf-f)} \\ &= &e^{\lambda t(Pf-f)}e^{\lambda s(Pf-f)} \\ &=& P_t(f)P_s(f) \dots \end{eqnarray}$$ which is a product and not a composition... Where is the catch?