Why are the two bonds in sulphur dioxide identical?
There are two ways to look at the $\ce{SO2}$ molecule. The valence bond way is this:

$\ce{SO2}$ is a bent molecule with C2v symmetry point
group. A valence bond theory approach considering just s and p
orbitals describes the bonding in terms of resonance between two
resonance structures:

The sulfur–oxygen bond has a bond order of 1.5 and does not invoke d
orbital participation. In terms of electron-counting formalism, the
sulfur atom has an oxidation state of +4 and a formal charge of +1.
(Wikipedia)

However, there is a molecular orbital approach which seems to give a deeper appreciation of the bonding. The picture is this:

(Picture source with detail explanation)
This is my simple explanation:

The sulfur is hybridized sp$^2$-p. The bond angle is 119$^o$ - you
can hardly get closer to 120$^o$!

We will assume that the oxygens are also hybridized sp$^2$-p.


Then the sigma bonding is designated in the diagram by the numbers 1 and 2 for the sigma bonds between sulfur and oxygen. Lone pairs are designated by numbers 3 (for S), and 4, 5, 6 and 7 (for O).
The pi system has three molecular orbitals, one from each atom and four electrons: one from each oxygen, two from sulfur. The lobes in these orbitals are drawn at the lower right side of the picture. Orbital 8 has the lowest energy, with all lobes bonding. The non-bonding pi orbital (#8) has reversed signs for the lobes of oxygens and therefore a node at sulfur. The antibonding pi orbital (#10) has two nodes and is unoccupied.
So the final count is two filled sigma bonds and one filled pi bond; five lone pairs and one filled non-bonding pi orbital. The formal bond order is 1.5, same as in the valence bond explanation with resonance.
The only thing I might change in the diagram is the relative energies of the sulfur orbitals and the oxygen orbitals: I would think they might be closed to the same energy. The diagram might be drawn as it is in order to make it clearer and give more room for all the numbers.
We just don't need d orbitals to explain anything, but their involvement cannot be entirely eliminated.