In fact, there are "nice" counterexamples. There is a notion of an inertial h-cobordism on $M$. It is an h-cobordism with both boundaries homeomorphic to $M$. By the s-cobordism theorem, all h-cobordisms are invertible in the sense that we can stack one on top of the other and get the identity cobordism.
Given any inertial h-cobordism we can extract one of the diagrams you suggest. Namely, stack the inverse h-cobordism on top of it and consider the embedding of $M$ into the middle. If the h-cobordism is nontrivial, than you get a counter example to your question. In fact, if $i:M \rightarrow M \times [0,1]$ is assumed to have a bicollar, these notions are exactly the same.
Page 3 of Whitehead torsion of inertial h-cobordisms talks about how to construct nontrivial examples.