Let $A$ be a bounded operator on a Banach space $X$. Consider the exponential function $x(t)=e^{tA}x:=\sum_{n=0}^{+\infty}\dfrac{t^nA^n}{n!}x$, for all $t\in \mathbb{R}$, where $x\in X$. If the function $t\mapsto x(t)=e^{tA}x$ is bounded in $\mathbb{R}$ then it is uniformly continuous. In fact $x(t)$ is a solution of the ordinary differential equation $$x'(t)=Ax(t).$$ It follows that $$x(t)=x(0)+\int_0^tAx(u)du.$$ Then $$\left|x(t)-x(s) \right|=\left|\int_s^tAx(u)du\right|\leq \left|A\right| \left|x\right|_\infty \left|t-s\right|.$$
So $x(t)$ is even Lipshitz.
We can visualize this intuitively in the case $X=\mathbb{R}$, the function $x(t)=e^{ta}x$ is bounded if and only if $a$ is pure imaginary, so $x(t)$ has the form $\cos(\omega t)+i\sin(\omega t)$ which is uniformly continuous.
My question concerns the case when $A$ is an unbounded operator which generates a strongly continuous group of operators $(T(t))_{t\in \mathbb{R}}$. We consider now the function $x(t)=T(t)x$. I tried to do the same manipulations as the function $t\mapsto e^{tA}x$. From some properties of strongly continuous groups we have $\int_0^tT(u)xdu\in D(A)$ for all $x\in X$, where $D(A)$ is the domain of $A$. In addition $$x(t)=T(t)x=x+A\int_0^tT(u)xdu=x+A\int_0^tx(u)du$$ Using this I tried $$\left|x(t)-x(s) \right|=\left|A\int_s^tx(u)du\right|.$$ But I stopped here since $A$ is not bounded. So my question is what can we say in this case ? is $x(t)=T(t)x$ uniformly continuous if it is bounded on $\mathbb{R}$ ?
Edit: I misunderstood the question. My answer does not adress it.
Consider e.g. the translation group on $L^2(\mathbb{R})$. These operators are isometries and thus the group is bounded, it is however not uniformly continuous.
This and more examples can be found on pp. 34 in the book of Engel and Nagel.