The answer is no in a more general setting of $(G,X)$-structures (see https://en.wikipedia.org/wiki/(G,X)-manifold).

A couple $(G,X)$ consists of a Lie group $G$ acting on a (simply connected) manifold $X$ transitively and analytically (i.e if $g\in G$ acts as identity on a (non-empty) open subset $U\subset X$ then it acts as identity globally).

Now, a (connected) manifold $M$ is said to admit a $(G,X)$-structure if it has an atlas of charts with values in $X$ such that all transitions are restrictions of elements of $G$ acting on $X$.
In this case, the universal cover $\widetilde{M}$ has a $(G,X)$-structure (by pulling back the one on $M$) and it admits a local diffeomorphism $D:\widetilde{M}\to X$ (called the developing map) which is constructed, roughly speaking, as analytically extending a local chart. If one proper charts $U\subset \widetilde{M}\to X$ is onto then there is no way to extend to a local diffeomorphism which contradicts the existence of the developing map.

In your example, $\mathbb{S}^1$ has a $(\operatorname{Aff}(\mathbb{R}), \mathbb{R})$-structure and if $U\to \mathbb{R}$ is a chart which is onto then it lifts to the universal cover to a chart which is also onto and this is impossible by the previous argument.