Is a cylinder a differentiable manifold?

If you want your cylinder to have an "edge" then intuitively, it should not be a differentiable manifold. If you take a point in a differentiable manifold, then there is a regularly parametrized curve through the point in each direction. (Just take a straight line in a chart in transport it onto the manifold.) This means that you can "pass through that point" in each direction on a smooth curve with non-zero velocity. If you apply this to a point on the edge, this means that you can get from the side of cylinder to the top of the cylinder by a smooth curve whose speed is non-vanishing at each point. But obviously this should not be possible at the edge, since the speed would have to jump from having non-zero vertical component to being horizontal. (What I am trying to explain in a non-technical way here is that the cylinder has no well defined tangent plane at points of the edge.)

The problem with a formal answer to your question is that in order to get a formal answer, you would first have to specify what it means to "make the cylinder into a manifold". As you said, you can choose a bijection to a sphere (which even is a homeomorphism) and use this to carry the manifold structure over to the cylinder. But in this way, you have "removed" or "flattened out" the edge. Certainly, the cylinder is not a submanifold in $\mathbb R^3$, which can be seen by making the above argument with tangent spaces precise.