Suppose that a set of sentences of a 1st order language has an infinite model $M$.


Under what conditions is there is a proper class-sized elementary extension of $M$?


How does the answer change if we begin with a *proper class* of sentences?



When does an infinite model have a proper class-sized elementary extension?
The answer to your main question is that in ZFC there is always such a proper-class elementary-extension. 


**Theorem.** In ZFC, every set-sized model in a set-sized
first-order language has a proper-class elementary extension.


**Proof.** This is easiest to see in the case that the global
axiom of choice holds, in other words if there is a class
well-ordering of the universe. So let me first explain that case.
Fix the global well-order and consider any fixed model $M\_0$ in a
set-sized first-order language. Using the upward
Löwenheim-Skolem theorem, there is a proper elementary
end-extension of $M\_0$, and we may let $M\_1$ be the least such
model arising in the well-order. Continuing transfinitely, picking the least elementary-extension at each stage and unions at limit stages, we may
build up an elementary chain $$M\_0\prec M\_1\prec\cdots\prec
M\_\alpha\prec\cdots$$ by a definable procedure whose union will be a proper-class
elementary extension of each of them and in particular of $M\_0$,
as desired.


But my next observation is that in ZFC you don't actually need
global choice. If we fix $M\_0$, then by the axiom of choice, we
may code $M\_0$ by a set of ordinals $A$. Consider the inner model
$L[A]$, which satisfies ZFC and global choice. Since $A$ codes
$M\_0$, we may undertake the argument of the previous paragraph
inside $L[A]$ to get a proper-class elementary extension of $M\_0$.
In the original universe $V$, then, we get an $A$-definable proper
class elementary-extension of $M\_0$, as desired. **QED**


Your second question, however, can fail in some models. I claim that it is possible to have a
definable proper-class-sized theory $T$ in a model of ZFC, such
that every subset of $T$ has a model, and so in particular the
theory is consistent, but there is no definable (allowing
parameters) model of all of $T$. For example, assume we are
working in a model of ZFC in which there is no definable linear
order. (I explained how to construct such a model in my answer to
Asaf Karagila's question,
[Does ZFC prove the universe is linearly orderable?](https://mathoverflow.net/a/110823/1946).) Let $T$ be the
theory of a linear order $<$, with a constant symbol $\hat a$ for
every object $a$. Every restriction of $T$ to only set-many
constants will have a model, since in ZFC every set is linearly
orderable, but in this model there is no definable model of all of
$T$, since there is no definable linear ordering of the universe.


