Consider the Banach space $L^1(\mathbb{R})$ of integrable functions $f:\mathbb{R}\to \mathbb{R}$. Consider the unbounded operator $A$ defined by $$(Af)(x)=f'(x)+a(x)f(x), \ \ \ x\in \mathbb{R}$$ for each absolutely continuous function $f$ such that $f'\in L^1(\mathbb{R})$, where $a:\mathbb{R}\to \mathbb{R}$ is a bounded function.
I wonder whether the $C_0-group$ generated by $A$ has the following explicit form $$(T(t)f)(x)=e^{a(x)t}f(t+x),\ \ \ \ \ t\in \mathbb{R}, \ \ \ x\in \mathbb{R}.$$ I am also interested in the case the function $a(.)$ is not bounded.
The operator $Af=(\frac{d}{dx}+a)f$ is, in a classical sense, similar to the differentiation operator because $$ Af = e^{-\int a(x)dx}\frac{d}{dx}e^{\int a(x)dx}f = M^{-1}\frac{d}{dx}Mf. $$ Because of this, any function of the operator $A$ is a corresponding function of $\frac{d}{dx}$. In particular, $$ e^{tA}f = e^{-\int a(x)dx}e^{t\frac{d}{dx}}e^{\int a(x)dx}f $$ Without regard to rigor (at least for the moment,) and assuming a Taylor series expansion for $g$, $$ e^{t\frac{d}{dx}}g = \sum_{n=0}^{\infty}\frac{t^{n}g^{(n)}(x)}{n!} = g(x+t). $$ That's enough to suggest the form of the group, if there is one: $$ \begin{align} (T(t)f)(x) & = e^{-\int_{0}^{x}a(y)dy}(e^{\int_{0}^{x+t}a(y)dy}f(x+t)) \\ & = e^{\int_{x}^{x+t}a(y)dy}f(x+t). \end{align} $$ Check the Answer: The above formula for $T(t)$ seems to work; I'll let you work out the group property. As for the the generator ... $$ \frac{d}{dt} T(t)f = \frac{d}{dt}\left[e^{\int_{x}^{x+t}a(y)dy}f(x+t)\right]\\ = e^{\int_{x}^{x+t}a(y)dy}f'(x+t)+e^{\int_{x}^{x+t}a(y)dy}a(x+t)f(x+t). $$ Setting $t=0$ gives $$ \frac{d}{dt}T(t)f|_{t=0} = f'(x)+a(x)f(x). $$ Of course, that's not a fully rigorous proof of the generator because you must carry out the calculation in the norm of $L^{1}$, but I'll leave it to you for the case where $f,f' \in L^{1}(\mathbb{R})$.
The expression for $T(t)$ defines a bounded operator on $L^{1}$; if $a$ is uniformly bounded by $M$, then you have the uniform bound $|\int_{x}^{x+t}a(y)dy| \le Mt$. Hence, $T(t)f \in L^{1}$ if $f\in L^{1}$, and $$ |(T(t)f)(x)| \le e^{tM}|f(x+t)|, \\ \|T(t)f\|_{L^{1}} \le e^{tM}\|f\|_{L^{1}}. $$ The $C^{0}$ property for $T$ is easy to verify: $$ \begin{align} \|T(t)f-f\|_{L^{1}} & =\left|\int_{-\infty}^{\infty}e^{\int_{x}^{x+t}a(y)dy}f(x+t)-f(x)dx\right| \\ & =\left|\int_{-\infty}^{\infty}(e^{\int_{x-t}^{x}a(y)dy}-1)f(x)dx\right| \\ & \le \int_{-\infty}^{\infty}|e^{\int_{x-t}^{x}a(y)dy}-1||f(x)|dx. \end{align} $$ The right side tends to $0$ as $t\rightarrow 0$ by the Lebesgue dominated convergence theorem.