Non-homeomorphic spaces such that taking away a point makes them homeomorphic

It is known that the Cantor set minus a point is (up to homeomorphisms) the unique zero-dimensional separable metric space without isolated points that is locally compact and not compact. In particular, the Cantor set minus a point is homeomorphic to the Cantor set minus two points (or minus any finite number of points).

Now you can take as $X$ the Cantor set and as $Y$ the Cantor set minus a point.

See also this math.stackexchange question.

If $X$ and $Y$ are Hausdorff then the one easily sees that the homeomorphism can not be induced by $\varphi$. But you did not request that.

Let $C_0$ denote a disjoint union of countably many copies of the Cantor set. It is easily seen that the Cantor set $C$ is homeomorphic to the one point compactification of $C_0$. Since $C$ is homogeneous, by removing any point from $C$ we get $C_0$. Then $C_0$ less any point is again homeomorphic to $C_0$. Thus take $X\cong C$, $Y\cong C_0$ and $\varphi$ to be any bijection between them.

For a minimal example consider the following two 3-element posets equipped with the Alexandroff topology (i.e. open sets are down-sets):

(a)   (b)                      (C)
  \   /                        / \
   \ /                        /   \
   (c)                      (A)   (B)

The bijection is given by: $a\mapsto A$, $b\mapsto B$, $c\mapsto C$.

The reconstruction conjecture for ordered sets (see e.g. Jean-Xavier Rampon, What is reconstruction for ordered sets?) - together with equivalence of finite posets and finite $T_0$ topological spaces (via the specialization order/Alexandroff topology functors) - says the are no such examples among finite $T_0$ spaces with at least four elements. However, the conjecture is open.

(The related reconstruction conjecture for graphs is better known and also open.)