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Is there a simple example of two spaces $A,B$ so that $A\not\cong B$, but which have the same cone $L(A)\cong L(B)$?

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    Let $A$ be a point and let $B$ be two discrete points. – 2017-02-19
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    Well, that is very simple. But somehow unfulfilling. – 2017-02-19

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Just take A =[0,1] and B= circle, both of them have the cone homeomorphic to closed disk.

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The whitehead manifold and $\mathbb R^3$ are not homeomorphic but their cartesian product with $\mathbb R$ are same and is $\mathbb R^4$ ie. the open cylinder on both of them are homeomorphic without the original spaces being homeomorphic.

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Surely yes. Take a look at the circle and cylinder, there are not homeomorphic to each other (but are homotopy equivalent) and their cone is just the ordinary cone as you know it from geometry.

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    $L(S^1\times(0,1) )$ has points with neighbourhoods homoemorphic to $\Bbb R^3$, but $L(S^1)$ is a $2$ manifold. They cannot be homoemorphic. Indeed I think $L(S^1\times(0,1) ) $ is $D^3$ whereas $L(S^1)$ is $D^2$ – 2017-02-19
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    You are right, $C(S^1 \times [0,1])$ is $D^3$, not $D^2$, sorry for the wrong suggestion. – 2017-02-19