$\DeclareMathOperator\R{R}\DeclareMathOperator\Z{Z}$No.

Indeed affine circles have been classified in the 50s by Kuiper. These are

1. the standard circle $C_1=\R/\Z$ (complete)
2. exotic circles $C_t=\R_{>0}/\langle t\rangle$ (non-complete) for $t>1$.

Any interval inside one of these circles is affinely equivalent to a bounded interval in the real line, hence to the interval $\mathopen]0,\mathclose 1[$, which is not affinely equivalent to $\R$ (e.g., because the automorphism group of the former is cyclic of order 2 when that of the latter is infinite — or because the former is non-complete and the second is complete).

$\DeclareMathOperator\R{R}\DeclareMathOperator\Z{Z}$No.

Indeed affine circles have been classified in the 50s by Kuiper. These are

1. the standard circle $C_1=\R/\Z$ (complete)
2. exotic circles $C_t=\R_{>0}/\langle t\rangle$ (non-complete) for $t>1$.

Any interval inside one of these circles is affinely equivalent to a bounded interval in the real line, hence to the interval $\mathopen]0,\mathclose 1[$, which is not affinely equivalent to $\R$ (e.g., because the automorphism group of the former is cyclic of order 2 when that of the latter is infinite — or because the former is non-complete and the second is complete).

Here's a plain classification-free argument. I claim that if a connected affine 1-manifold $X$ has an open subset $U$ isomorphic to $\R$, then $X=U$.

Proof: otherwise, let $x$ be an element of $\bar{U}-U$. Let $J$ be a small enough interval around $x$ and $u:\mathopen]-1,1\mathclose[\to J$ mapping $0$ to $x$. Let $v:\R\to U$ be an affine isomorphism, with $x=\lim_{+\infty}v$. So the partially defined map $w=u^{-1}\circ v$ is affine, and injective and well-defined near $+\infty$, such that $\lim_{+\infty}w=0$. But near infinity, a bounded affine map is constant. We get a contradiction.