As long as the set of sentences is given by any "reasonable" grammar, it will be decidable. The same proof as usual can be modified: you just provide a parsing algorithm for the language. It's sufficient to show that the language has a context-free grammar, which it will even if you add more function symbols. It doesn't matter what the symbols mean; all you are asking about is whether it is possible to tell whether some string of symbols is actually a formula.
The set of logically valid formulas for first-order logic is not decidable, however. It is only computably enumerable, and that follows from the completeness theorem for first-order logic. So if you added new symbols with different semantics, in a way that made the completeness theorem fail, then the set of logically valid formulas in your new language might not be computably enumerable. This is the case for second order logic, where the set of logically valid formulas (in full second-order semantics) is very far from being computably enumerable.