 Solution

No. The parameter-free partial typeis finitely satisfiable in the ordered real field: choose a number larger than the finitely many bounds. By compactness theorem it extends to a complete type over the empty set. But no real number realizes it, by the Archimedean property. Therefore the ordered real field is not an omega-saturated model.

The named unit and addition make every integer bound expressible, even without extra parameters. This is why the obstruction is present in the ordered-field language and absent in the two reducts above.