Build form elements in formBuild() only, not in ajax callback. The ajax callback is only for preparing the ajax response. When the ajax callback is called, formBuild() has already run with the updated $form.
Your ajax response is empty. You didn't add any ajax command to it. See ajax examples how you can do this.
But in your case you don't need any code in ajax callback. It would be enough to simply return the changed part of $form in a div container.