You are right that the classification does not imply that $C$ and $[a, b]$ are diffeomorphic. It may very well be that $C$ is diffeomorphic to a circle, and that $c$ winds the interval $[a, b]$ 3.5 times around it. (The minute hand of the clock was giving you a hint.)
But assuming that $C$ has nonempty boundary, it follows indeed that the immersion $c$ is injective (i.e. is an embedding), as shown in a linked question:
Is a smooth immersion $c: [a,b] \to M$ injective if its image is a 1-manifold with non-empty boundary?
In particular, the boundary of $C$ has cardinality $2$ and equals $\{ c(a), c(b) \}$.
Well, manifolds with corners are not one and the same as topological manifolds. Let me come back to that point later, and address the easy parts of your question first.
In fact $\partial C=(S^{1}\times [0,1])\cup (D^{2}\times\{0,1\})$ is indeed homeomorphic to $S^2$. One way to see this is to subdivide $S^2$ at two latitude lines: the arctic circle at $67^\circ$ north latitude and antarctic circle at $67^\circ$ south latitude. The required homeomorphism takes the northern polar cap above $67^\circ$ north latitude to $D^2 \times \{1\}$; it takes the southern polar cap to $D^2 \times \{0\}$; and it takes everything between the arctic and antarctic circles to $S^1 \times [0,1]$. You can write down a completely concrete formula for this homeomorphism using a piecewise smooth formula with three pieces, expressed in terms of spherical coordinates on $S^2$ and polar coordinates on $D^2$.
However, this does not give a manifold-with-corners structure to $S^2$: the "corner structure" might appear to be the arctic and antarctic circles; but those are circles, they are disjoint from the topological boundary (which is actually empty), and the only "corners" on a 2-manifold-with-corners are isolated points on the topological boundary.
In fact the only manifold-with-corners structure on $S^2$ is one whose charts have values in $(0,\infty)^2$, which is equivalent to an actual topological manifold structure (with empty boundary).
As for your prism structure, you simply miscounted the edges, there are 9 edges: three around the arctic circle; three cutting across from the arctic circle to the antarctic circle; and three around the antarctic circle.
So what about your one and the same comment? It helps here to think in terms of category theory. Intuitively, a manifold with corners has more structure than a topological manifold. By forgetting this structure, you get a forgetful functor from the category of manifolds with corners to the category of topological manifolds.
What exactly has been forgotten? You write that the model space $[0,\infty)^d$ for a manifold with corners is homeomorphic to the model space $[0,\infty) \times \mathbb R^{d-1}$. This is true. However, the definition of manifold with corners has a rather strong restriction on overlap maps, which you have not mentioned. If the $U,V \subset [0,\infty)^d$ are the open subsets that are the targets of two charts in your manifold-with-corners atlas, and if $\psi : U \to V$ is the overlap map itself, then not only must $\psi$ be a homeomorphism (from $U$ to $V$), but $\psi$ must respect the corner structures. I suggest that you read exactly what atlases are, focussing on the overlap condition, in whatever textbook you might have on this topic. But in brief: the map $\psi$ must take the origin in $U$ to the origin in $V$ (assuming $U$ or $V$ contains the origin); it must take the union of the coordinate axes (intersected with $U$) to the union of the coordinate axes (intersected with $V$); it must take the union of the coordinate 2-planes to the union of the coordinate 2-planes; and so on. All of this extra structure is forgotten when you pass to the category of topological manifolds.
Best Answer
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 $X$.