Yes, your desired equality is true: regarding the left regular representation as
$$
\operatorname{Ind}_{\{0\}}^{\mathbf R}1,
$$
it becomes a special case of the characterization of smooth vectors in induced representations by N. S. Poulsen, [*On $C^\infty$-vectors and intertwining bilinear forms for representations of Lie groups*][1]. J. Functional Analysis **9** (1972), 87–120, Theorem 5.1.

**Edit** to clear up your extra question: For $G=\mathbf R$, we have $\mathfrak g=\mathbf R$ and $\exp$ is just the identity $\mathbf R\to\mathbf R$. So Poulsen's second displayed formula on p. 113 says
\begin{align}
(Xf)(x)
&=\Bigl.\frac d{dt}f(\exp(-tX)\cdot x)\Bigr|_{t=0}\\
&=\Bigl.\frac d{dt}f(x-tX)\Bigr|_{t=0}= - Xf'(x).
\end{align}
Fixing $X=-1$ (basis of $\mathfrak g$) we get $X^\alpha f=f^{(\alpha)}$ and so Poulsen's first displayed formula on p. 114,
$\mathbf D_\infty(U_2)=\{f\in C^\infty(G)\mid X^\alpha f\in L^2(G) \text{ for all } \alpha\}$, is exactly your desired equality. 

  [1]: http://dx.doi.org/10.1016/0022-1236(72)90016-X