I am trying 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 $f\in \mathcal{C}_0(\mathbb{R}^d)$, $P$ being a probability kernel), is a strongly-continuous contraction semigroup (on $(\mathcal{C}_0(\mathbb{R}^d),||\, .\, ||_{\infty})$).
It remains for me to prove that it is a contraction : $||P_t(f)||_{\infty}\leq || f||_{\infty}$.
At best I could find that $$||P_t(f)||_{\infty}\leq e^{|\lambda t | ||P-I ||}||f||_{\infty} $$ and $||P-I||\leq 2$, wich is not sufficient... (since, of course, in all case $e^{|\lambda t | ||P-I ||}\geq 1$)
Secondly, I need to prove the positiveness of the operator ($P_t(f)\geq 0$ if $f\geq 0$), and I have no clue for that...
There is the following general statement:
Since any operator commutes with the identity operator, we find
$$\begin{align*} P_t f &= e^{\lambda t(P-I)} f= e^{-\lambda t I} e^{\lambda tP}f = e^{-\lambda t} e^{\lambda t P} f. \tag{2} \end{align*}$$
This identity allows us to prove the desired properties of the semigroup. $P$ is a positive operator and therefore $P^n$ is a operator for any $n \geq 1$. Thus,
$$P_t f \stackrel{(2)}{=} e^{-\lambda t} \sum_{n \geq 0} \frac{(\lambda t)^n}{n!} \underbrace{P^n f}_{\geq 0} \geq 0$$
for any $f \geq 0$. This shows the positivity of the semigroup. To prove that $P_t$ is a contraction we note that
$$\|e^{\lambda t P} f\|_{\infty} \leq \sum_{n \geq 0} \frac{|\lambda t|^n}{n!} \|P\|^n \|f\|_{\infty} = e^{\lambda t \|P\|} \|f\|_{\infty}$$
(this is essentially the inequality you mentioned in the solution attempt in your question). Since $\|P\|=1$, we get
$$\|e^{\lambda tP} f\|_{\infty} \leq e^{\lambda t}\|f\|_{\infty}.\tag{3}$$
Thus, by $(2)$,
$$\|P_t f\|_{\infty} = e^{-\lambda t} \|e^{\lambda tP}f\|_{\infty} \stackrel{(3)}{\leq} \|f\|_{\infty}.$$