The actual full question: If there are free variables in the premises of a natural deduction that doesn't lead to contradiction, then is there a non-empty domain such that its constant terms satisfy the conclusion?
Well, let's consider a set of premises such that it has free variables somewhere. Using the rules of inference, I can get to a conclusion that also has the free variables. Since these free variables are any arbitrary terms, there has to be a non-empty domain that satisfy it. Is it correct? Does the domain have to be previously defined or can it be arbitrary?
Also, is the following inference ok? $$Px \vdash \exists x (Px)$$
Yes; in Natural Deduction the above inference is a correct application of the $(\exists \text I)$ rule.
The standard semantics for classical predicate logic assume that the domain of interpretation is not-empty.
Thus, the above inference is sound because in every domain where $P(x)$ holds, it is true that there is some element that is $P$.
What does it mean that a formula with a free variables is true ?
Two possibilities : either that (i) $P(x)$ is a shorthand for $\forall x P(x)$, or (ii) there is an assignment function $s : \text {Var} \to A$, where $A$ is the domain of the interpretation $\mathfrak A$, such that $\mathfrak A \vDash P(x)[s]$.
In both cases, it is not necessary that the language has constant symbols.