Obviously, we have $P_0 = 1$. Let $s,t\ge 0$. Then using the fact that the operators $P_j$ commute and $P_j^2 = P_j$ for all $j$, we obtain \begin{align*} P_sP_t &= \prod_{j=1}^m \Big[(P_j + e^{-s}(1 - P_j))(P_j + e^{-t}(1 - P_j))\Big] \\ &= \prod_{j=1}^m (P_j^2 + e^{-s}(P_j - P_j^2) + e^{-t}(P_j - P_j^2) + e^{-(s+t)}(1 - 2P_j + P_j^2))\\ &= \prod_{j=1}^m (P_j + 0 + 0 + e^{-(s+t)}(1 - P_j))\\ &= P_{s+t}. \end{align*}

Since $\|P_j\|\le 1$, we have for all $t\ge 0$: \[ \|P_t\|\le \prod_{j=1}^m \|e^{-t} + (1-e^{-t})P_j\| \le \prod_{j=1}^m (e^{-t} + (1-e^{-t})) = 1. \] Thus, $P_t$ is a contraction semigroup on $E$.

For each $j$, the function $t\mapsto P_j + e^{-t}(1-P_j)$ is continuous as a mapping between $\R^+_0$ and the space of linear bounded operators on $E$, equipped with the operator norm, hence the function $t\mapsto P_t$ is also continuous. In particular, this implies that for all $x\in E$ the function $t\mapsto P_tx$ is continuous, so $P_t$ is a continuous contraction semigroup.