Yes, your desired equality is true: regarding the left regular representation as $$ \operatorname{Ind}_{\{0\}}^{\mathbb R}1, $$$$ \operatorname{Ind}_{\{0\}}^{\mathbf R}1, $$ 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. 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.