I saw that it's written in a couple of places, but couldn't find proper proof for this.
I saw it for example in:
I would like to find a reference to a proof of this (as I'm sure some book or a thing like that addresses the problem) or the proof itself.
I hope it's legit to ask for this.
Let $E$ be a vector space over the field $\Bbb K$, and let
$$ \mathcal P(E) = \big\{F\subseteq E\;\; : \;\; F \text{ subspace of } E, \;\dim(F)=1\big\} \;=\; \big\{\text{vector (straight) lines of } E\big\} $$
be the projective space induced by $E$.
If $H$ is a hyperplane of $E$, i.e. $\mathcal P(H)\:$ is, by definition, a hyperplane of $\mathcal P(E)$, we want to prove that
$$ \mathcal A := \mathcal P(E)\setminus\mathcal P(H) \tag{1}$$
is an affine space contained in $\mathcal P(E)$, with associated translation space $H$, also denoted $\;\mathcal A(H)$.
Since $H$ is a hyperplane of $E$, we have $\;\text{codim}(H)=1$, hence there exists $\;a\in E, \;a\neq0,$ such that
$$ E=H\oplus \langle a\rangle. \tag{2}$$
For each $v\in E$, we have $\;v=h+\lambda a\;$ for suitable and uniquely determined $\;h\in H\;$ and $\;\lambda\in\Bbb K$; as the vectors $\;h\in H\;$ are obtained for $\;\lambda= 0$, we can write
$$ \mathcal P(E) = \mathcal P(H)\,\dot\cup\,\big\{\big\langle h+\lambda a\big\rangle\;\; :\;\;h\in H, \;\lambda\in\Bbb R, \;\lambda\neq 0\big\}, \tag{3}$$
where $\;\dot\cup\;$ denote the disjoint union.
Now, we observe that the set $\,\mathcal A\,$ after the disjoint union sign in $(3)$ is equal to
$$ \mathcal A = \big\{\big\langle h+a\big\rangle\;\;:\;\;h\in H\big\}. $$
Indeed, being $\;\lambda\neq 0,\;$ we have
$$ \big\langle h+\lambda a\big\rangle = \big\langle \lambda\lambda^{-1}h+\lambda a\big\rangle = \big\langle \lambda\big(\lambda^{-1}h+a\big)\big\rangle = \big\langle\lambda^{-1}h+a\big\rangle, $$
and obviously $\;\lambda^{-1}h\;$ describes $\;H\;$ as $\;h\;$ describes $\;H$.
The points of $\;\mathcal A(H)\;$ are therefore by definition the projective points lying in $\; \mathcal P(E)\setminus \mathcal P(H)\;$, i.e. $(1)$ holds.
Finally, $\;\mathcal A(H)\;$ is an affine space through the structure defined by the function
$$ \phi\; : \; \mathcal A(H) \times \mathcal A(H) \;\;\longrightarrow\;\; H $$
defined by
$$ \phi\big(\big\langle h_1+a\big\rangle, \big\langle h_2+a\big\rangle\big) := h_2-h_1, $$
as it is easy to verify.
As a last observation, it can also be easily verified that the preceding construction of $\;\mathcal A(H)\;$ does not depend on the choice of $\;a\;$ in $(2)$. In fact, if $\,a'\neq0\,$ is another vector satisfying $(2)$, and denoted $\,\mathcal A'(H)\,$ the affine space so obtained, the function
$$ f\;:\; \langle h+a\rangle \longmapsto \langle h+a'\rangle\; $$
from $\;\mathcal A(H)\;$ to $\;\mathcal A'(H)\;$ is an affine isomorphism.