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.
If the ball is embedded nicely enough, it will have a neighborhood which is contractible (ambiently contractible, even). Let me pretend each ball is in a single coordinate chart. If the manifold is connected, take a path connecting the balls, which we can take to be embedded. Take a small normal neighborhood.
So we have two balls with a little tube connecting them. By the usual look-at-the-complement-of-a-slightly-smaller-neighborhood, the manifold is this, plus stuff glued to the boundary. But then it's obvious that there is a homeomorphism which swaps the two balls, by going through the tube.
Best Answer
Hint: Verify that interiors of Mazur manifolds are not simply-connected at infinity unlike $R^4$. use the fact that the homology spheres bounding Mazur manifolds are not simply-connected.
Here are some details. Let's start with the definition of (topological) Mazur manifolds. Let $W$ be a compact contractible 4-dimensional manifold $W$ (necessarily with nonempty boundary $Y=\partial W$). A manifold $W$ is called a topological Mazur manifold if the boundary $Y$ is not simply-connected. Equivalently (this part is the 3-dimensional Poincare conjecture), $Y$ is not homeomorphic to the sphere $S^3$.
Here is a bit you can ignore if you do not know what homology means:
The above definition can be "reversed": Let $Y$ be a closed (compact and with empty boundary) 3-dimensional homology sphere, i.e. a connected nonempty orientable 3-dimensional manifold such that $H_1(Y)=0$. Then it is proven by Michael Freedman that for every manifold $Y$ with this property there exists a contractible compact 4-dimensional manifold $W$ (with boundary) such that $\partial W$ is homeomorphic to $Y$. Moreover, topology of $Y$ uniquely determines topology of $W$ (this is due to Richard Stong, Freedman's student). Thus, whenever $Y$ is not simply-connected, the manifold $W$ is a topological Mazur manifold.
Now, you are saying:
This part is not quite right: Additionally, a Mazur manifold is required to be compact. Also, the fact that such a manifold is not homeomorphic to $E^4={\mathbb R}^4$ is quite easy: A compact topological space cannot be homeomorphic to a noncompact one. Furthermore, a manifold with nonempty boundary cannot be homeomorphic to a manifold with empty boundary, such as ${\mathbb R}^4$.
I need one more definition:
Definition. A topological space $X$ is said to be simply-connected at infinity if the following two properties hold:
a. $X$ is 1-ended, meaning that for every compact $K\subset X$ there exists a compact $C\subset X$ whose interior contains $K$, such that any two points $x, y\notin C$ can be connected by a path in $X\setminus K$. (One can think of this property as "path-connected at infinity".)
b. For every compact $K\subset X$ there exists a compact $C\subset X$ whose interior contains $K$, such that every continuous map $f: S^1\to X\setminus C$ extends to a continuous map $F: D^2\to X\setminus K$.
Lemma 1. For each $n\ge 3$, the Euclidean space $E^n$ is simply-connected at infinity.
Proof. Each compact $K\subset E^n$ is contained in a round ball $B\subset E^n$. Take $C:= B$ for both parts (a) and (b) of the definition. Then $E^n\setminus B$ is homeomorphic to $S^{n-1}\times (0,\infty)$ (this is proven using spherical coordinates). Since $n\ge 2$, $S^{n-1}$ is simply-connected and, therefore, the product $S^{n-1}\times (0,\infty)$ is also simply-connected. hence, every map $S^1\to E^n\setminus B$ extends to a map $D^2\to E^n\setminus B$. Lemma follows. qed
Lemma 2. Suppose that $W$ is a compact manifold with connected nonempty boundary $Y=\partial W$ and $Y$ is not simply-connected. Then the manifold $X:= W\setminus Y$ is not simply-connected at infinity.
Proof. It is known that the boundary of $W$ has collar, which is an open neighborhood $U$ of $Y$ in $W$ homeomorphic to $Y\times [0,1)$.
See also Proposition 3.42 in Hatcher's "Algebraic Topology" for a self-contained proof.
In what follows, I will identity points of $U$ with their images in $Y\times [0,1)$.
Pick a homotopically-nontrivial loop $c_0: S^1\to Y$, i.e. one which does not extend to a map $D^2\to Y$. Then for every $t\in (0,1)$, the loop $$ c_t: S^1\to Y\times \{t\}\subset U, c_t(s)=(t, c_0(s)) $$ is homotopically nontrivial in $U$. (This is because the inclusion map $Y\to U$ is a homotopy-equivalence.)
Now, take the compact $K:= W\setminus U$. For every compact $C\subset X=int(W)= W\setminus Y$, there exists $t\in (0,1)$ such that $$ C\subset K\cup Y\times (t,1). $$ Therefore, there exists a loop $c_t$ as above, whose image is in $W\setminus C$ and such that $c_t$ is homotopically nontrivial in $U$. Hence, $X$ is not simply-connected at infinity.
Corollary. The interior of each topological Mazur manifold $W$ is not homeomorphic to $E^4$.
Proof. If $X=int(W)$ is homeomorphic to $E^4$, then $X$ is simply-connected at infinity, contradicting Lemma 2. qed