Let $M, N$ be smooth manifolds such that $M \times \mathbb{R}$ is diffeomorphic to $N \times \mathbb{R}$. Under which extra conditions is it possible to construct a diffeomorphism $f : M \times \mathbb{R} \rightarrow N \times \mathbb{R}$ that maps $M \times 0$ into $N \times (-r, r)$ for some $r > 0$ ?
Note that this would produce a proper $h$-cobordism between $M$ and $N$ by taking the region between $f(M \times 0)$ and $N \times r$ in $N \times \mathbb{R}$. If we assume that the fundamental group of $M$ has vanishing Whitehead group (e.g. if $M$ is simply connected or $\pi_1 M$ f.g. free), Siebenmann's proper $s$-cobordism theorem applies and shows that $M$ and $N$ are actually diffeomorphic.
Some obvious answers to my question are when $M$ is compact or $M$ and $N$ are a priori diffeomorphic.
I am particularly interested in the case where $M$ is homeomorphic to $N = \mathbb{R}^5$. I know that $\mathbb{R}^5$ has a unique smooth structure, the point is that this is what I want to conclude by the above argument while assuming that $\mathbb{R}^6$ has a unique smooth structure.
Decoding for the $N = \mathbb{R}^n$ case: If $N = \mathbb{R}^n$, the question might be more accessible to geometric intuition, namely: Given a properly embedded codimension 1 submanifold $S \subset \mathbb{R}^{n+1}$, is there a self diffeomorphism $\mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1}$ that carries $S$ into $\mathbb{R}^{n} \times (-r, r)$ for some $ r > 0 $ ?
Edit: I turned the special question into a Math Overflow post which got answered.
Be careful with how you apply the Proper s-cobordism Theorem. The obstruction to an inclusion being an infinite simple homotopy equivalence is more complicated than you indicate. (See Section 3 of Siebenmann's "Infinite Simple Homotopy Types".) It involves not just the fundamental group of the space, but also its fundamenatal group system at infinity. In particular, it is not enough to know that the Whitehead group of the fundamental group of M is trivial. In the special case of R^n you will be okay since R^n is simply connected at infinity (n>2).