Let $S(x)$ be a set of $\mathcal{L}$-formulas (containing at most the free variable $x$). Is there an elementary extension $\mathcal{N}$ of $\mathcal{M}$ such that $\mathcal{N} \models Th(\mathcal{M}) \wedge S(c)$, if $Th(\mathcal{M}) \cup S(x)$ is consistent?
I think that this is an obvious conclusion but I cannot find a rigorous way to construct $\mathcal{N}$ from an arbitrary model of $Th(\mathcal{M}) \cup S$, say $\mathcal{A}$. Do I need to start with a construction of an elementary map from $\mathcal{M}$ to $\mathcal{A}$?
Thanks in advance.
I think you ask for some $N\succeq M$ that realizes $S(x)$, that is, that contains an element $a$ such that $N\models S(a)$.
The answer is `yes', if $S(x)$ is finitely consistent in $M$.
Let $L'$ be the language that contains $M$ as a set of constants plus a new constant $c$. Let $T'={\rm Th}(M)\cup S(c)$. This is finitely consistent (every finite subset of it holds in the expansion of $M$ that interprets $c$ in the obvious way). By compactness $T'$ has a model $N'$.
The required $N$ is the reduced of $N'$ to the original language $L$. The element of $N$ that realizes $S(x)$ is the element that in $N'$ interprets the constant $c$.