Is a smooth immersion $c: [a,b] \to M$ injective if its image is a 1-manifold with non-empty boundary?

Yes. The only $1$-manifolds, up to diffeomorphism, are $S^1, \mathbb R$, $[0, 1]$ and $[0, 1)$.

Fact. For $X$ $1$-dimensional and $Y$ $1$-dimensional, with $Y$ not diffeomorphic to $S^1$, a smooth immersion $c : X \to Y$ is injective.

Proof. Take a point $t \in X$ such that $c(t)$ equals some $c(s)$ with $s \neq t$. Because $c$ is an immersion, $c$ is approximately linear around $t$, so there exists $s \in X$ and a segment $[t, s] \subset X$ such that $c(s) = c(t)$ and such that $c$ is injective on $[t, s)$. Here, by segment, I mean a diffeomorphism on its image $\phi : [p, q] \to X$ with $\phi(p) = t$, $\phi(q) = s$. Now $c \circ \phi$ is a smooth immersion $[p, q] \to \mathbb R$ with $\phi(p) = \phi(q)$. By the mean value theorem, its derivative vanishes somewhere. A contradiction. $\square$

To prove that classification, here's a hint: consider a maximal chart $(0,1) \to X$. Use maximality to prove that its image misses at most $2$ points of $X$.