Is any manifold that can be parametrized with only one chart an affine space? From Wikipedia:
An affine space is a set A together with a vector space $\overrightarrow{A}$, and a transitive and free action of the additive group of $\overrightarrow{A}$ on the set A.
Now let's say I have a manifold that is completely covered by just one chart $\phi: M \rightarrow \overrightarrow{A}$.
I can easily define an operation "$+$" between a Member of $M$ and an element of $\overrightarrow{A}$ by:
$$
m + v = \phi^{-1}(\phi(a) + v)
$$
That satisfies all of the required properties:
$$
m + 0 = \phi^{-1}(\phi(m) + 0) = \phi^{-1}(\phi(m)) = m
(m+v)+w = \phi^{-1}(\phi(m+v) + w) = \phi^{-1}(\phi(\phi^{-1}(\phi(m) + v)) + w)  \\
= \phi^{-1}(\phi(m) + v + w) = m + (v + w)
$$
I have to require that $\phi: M \rightarrow \overrightarrow(A)$ is surjective, so this proof works at every point and for every v.
By this definition, a general manifold could without any problem be an affine space, I just need to provide a more sophisticated "$+$" operation. Is this valid? My Intuition says that it shouldn't be.
 A: Based on clarifications in the comments, let's write $(A, *)$ to connote an affine space, i.e., the underlying set $A$ of vectors together with the addititonal structure of displacements and affine scalings about points.
Now suppose $M \subset A$ has the same cardinality as $A$, and fix a bijection $\phi:M \to A$. The questions appear to be,

*

*Can't we view $M$ as an affine space literally, and

*Doesn't this flout common sense?

For the first, yes we can use the bijection $\phi$ to transfer the affine structure from $A$ to $M$, e.g., defining
$$
m +_{\phi} m' = \phi^{-1}(\phi(m) + \phi(m')),\quad m, m' \in M.
$$
This operation $+_{\phi}$ does literally comprise part of the structure of an affine space of $M$. It's also, to make a comical understatement, typically not equal to $+$. Chances are, the affine structure $*_{\phi}$ is so pathological it's uninteresting, analogous to obfuscated code, or to a pile of wood chips compared to a living tree.
To resolve the second, we need to examine "common sense" more carefully. When we breezily say "$M \subset A$ is an affine space," we mean "$M \subset A$ and $(M, *)$ is an affine space." In detail, "$(M, *)$ is an affine space" means $M$ is closed under affine translation and scaling as defined in $(A, *)$. In this sense, an open subset $M \subset A$ is an affine space if and only if $M = A$, which is presumably common sense.
Strictly, we cannot view a "bare set" such as $\mathbf{R}^{n}$ as being "a group" or "a vector space" or "a smooth manifold." In practice, the structure is usually understood (the additive group, the real vector space, the Euclidean manifold), and we get breezy. It's crucial to internalize, however, that these structures are accessories "sold separately." Different accessories (the rational vector space, the discrete topology) give the set different mathematical characteristics.
Regarding the final sentences of the question: An affine space (over the reals, say) can be viewed as a smooth manifold in a natural way, with the identity map as a global chart. A smooth manifold of positive dimension cannot, however, be naturally regarded as an affine space, even if $M$ is an open subset of an affine space diffeomorphic to the whole space. Loosely, this would be possible only if all diffeomorphisms $M \to A$ induced the same affine structure.
