I am supposed to provide a mechanism for the above transformation. I observed that the order of substituents on the product is same as that on furan, and since furan undergoes Diels–Alder reactions, I came up with the following mechanism:
Is my Diels–Alder product correct? After this, however, I have no idea how aromaticity is restored.
Your Diels–Alder adduct is pretty much correct. Without any further evidence we cannot definitely say whether the exo or the endo adduct is formed (you have drawn the exo adduct, which generally arises via thermodynamic control). However, the stereochemistry of this adduct does not affect the final product, so it does not matter which one is formed.
By the way, you also should think about the regioselectivity of the Diels-Alder reaction, i.e. why do the $\ce{NH2}$ and $\ce{COMe}$ end up ortho to each other and not meta. You need to have a better reason than simply "that's the way it was drawn in the product". Generally, this is rationalised by looking for the largest coefficients in the HOMO of the diene and the LUMO of the dienophile. The HOMO of 2-aminofuran is likely to have a large coefficient on C–5 (opposite the amino group), whereas the LUMO of methyl vinyl ketone has a large coefficient β to the carbonyl group.
The subsequent steps are not anything fancy. Generally, $\mathrm{sp^3}$ carbons that are bonded to two heteroatoms tend to collapse (e.g. tetrahedral intermediates in nucleophilic acyl substitution; hydrates; etc.) In this case the nitrogen kicks out the oxygen. Note that the other option of oxygen kicking out nitrogen is not possible, as that would lead to a bridgehead double bond.
Ordinarily you need some sort of acid or base catalyst, but presumably here, the thermodynamic driving force for the collapse is large (relief of ring strain + product is aromatic), and the high temperature provides the energy required to overcome the activation barrier. The last elimination step probably occurs via an E1 mechanism, since there's no good base present.
