Here is proposition 4.20 from Görtz and Wedhorn's Algebraic Geometry I.
It seems to me that the right square in (4.5.1) is completely unnecessary: We can always choose $S = X$ making the right square trivial (and replacing $X \times_S Y$ by just $Y$). This however doesn't loose any generality since we just relax the assumptions: instead of needing two Cartesian squares of schemes, we now only need one.
The right square also isn't mentioned in the results of the theorem.
Am I right? And if so what is the reason why the authors included the right square in the statement of the theorem?