Yes, in fact whenever $C$ is a non-empty Zariski open inside a smooth connected curve $C_1$, then the complement $C_1\setminus C$ is finite. This is easy to see in the case that $C$ is affine, since in that case $C_1=D(f)$ is the locus of non-vanishing of a non-zero algebraic function $f$ on $C_1$. So the complement $C_1\setminus C$ is the (reduced) vanishing locus of $f$, which has dimension $0$ and hence is finite.
The general case is easy to deduce from this, either by covering $C_1$ with open affines, or by removing a single closed point $c$ of $C_1$ and using that $C_1\setminus\{c\}$ is affine, as you remark.
Regarding your final question, $C(\mathbb A_k)\to C_1(\mathbb A_k)$ is always injective, but will not be bijective in general. For injectivity, you have the general fact that if $C$ is a Zariski-open subscheme of a scheme $C_1$, then a morphism of schemes $f\colon S\to C_1$ factors through $C$ if and only if the image of $f$ is contained in the open subset $C\subseteq C_1$. (I thought this fact was in Hartshorne, but couldn't find it on a quick search.) This says that the map $C(S)\rightarrow C_1(S)$ is injective in general.
Now the map $C(\mathbb A_k)\to C_1(\mathbb A_k)$ certainly can be bijective in degenerate cases. For example, if $C_1(k_v)=\emptyset$ for some place $v$ of $k$, then both sets are empty and so any map between them is a bijection.
But you shouldn't suppose that this map will be bijective in any generality. For example, if $C_1\setminus C$ contains a $k$-rational point $c$, then the map $C(S)\hookrightarrow C_1(S)$ is not bijective for any non-empty $k$-scheme $S$. Indeed, the morphism $f\colon S\to C_1$ collapsing all of $S$ to the point $c$ is an element of $C_1(S)$ that does not lie in $C(S)$, since the image of $f$ is not contained in $C$.
