Calculating Junction Temp of Switching Regulator
Your loss estimate \$(V_{in}-V_{out})*I_{out}\$ is for a linear regulator, not for a switcher, and even then it doesn’t account for \$I_{q}\$, the regulator internal current to power its own stuff.

The correct gross loss calculation is \$I_{in} V_{in} - I_{out} V_{out}\$. This is true for either a linear or a switcher, as it will capture \$I_{q}\$ in the gross figure.

Working from there, the losses break down between the chip itself and components external to the chip. A good portion of the loss is IR drop in the inductor and ESR loss in the filter capacitor.

Conveniently, WebBench has tabs with separate power dissipation figures for each major component. They use the IC's figure to calculate \$T_{j}\$, based on their estimate of overall thermal resistance to ambient (\$θ_{ja}\$) based on their reference layout.

Ultimately, the sum of all the power dissipation figures is the gross loss, which should match the figure given by \$I_{in} V_{in} - I_{out} V_{out}\$.