Disclaimer
This thread is just to record. See: Answer own Question
It is stated as question for jeopardy. Have fun. :)
Problem
Given a finite Borel measure $\mu(\mathbb{R})<\infty$ and a Banach space $E$.
Construct a strongly continuous group $T(t)$.
Find a trajectory that is not uniform limit: $$x(t):=T(t)x_0:\quad s_n\nrightarrow x$$ so that integrals must be not taken as uniform but as Bochner or Riemann.
Consider the Banach space $\ell^\infty(\mathbb{N})$.
Define the group on the basis: $$T(t)\chi_{\{n\}}:=e^{it/n!}\chi_{\{n\}}$$ and extended by linearity and continuity.
Then it is a group: $$T(s+t)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)=\sum_{k\in\mathbb{N}}e^{i(s+t)/k!}a_k\chi_{\{k\}}=T(t)T(s)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)$$ $$T(0)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)=\sum_{k\in\mathbb{N}}e^{i0/k!}a_k\chi_{\{k\}}=1\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)$$ of bounded even isometric operators: $$\left\|T(t)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)\right\|=\sup_{k\in\mathbb{N}}\left|e^{it/k!}\right|\cdot|a_k|=\sup_{k\in\mathbb{N}}1\cdot|a_k|=\left\|\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right\|$$ that is even uniformly continuous: $$\left\|(T(h)-T(0))\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)\right\|=\sup_{k\in\mathbb{N}}\left|e^{ih/k!}-1\right|\cdot|a_k|\leq\left|e^{ih/1!}-1\right|\cdot\left\|\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right\|$$ So it satisfies all requirements.
Consider a finite Borel measure $\mu(\mathbb{R})<\infty$.
Regard the trajectory: $$x_0:=\sum_{k\in\mathbb{N}}1\chi_{\{k\}}:\quad x(t):=T(t)x_0$$
Then its range is bounded but not precompact as: $$\|x((2M)!\pi)-x((2N)!\pi)\|=\sup_{k\in\mathbb{N}}\left|e^{i(2M)!\pi/k!}-e^{i(2N)!\pi/k!}\right|\geq\left|e^{i(2M)!\pi/(2N)!}-e^{i(2N)!\pi/(2N)!}\right|\stackrel{M<N}{=}2$$ so that it can't be uniform limit of simple functions $s_n\nrightarrow x$.