A theory $T$ is called *near model complete* if every formula is equivalent to a Boolean combination of existential formulas mod $T$. I wonder whether there is an equivalent "semantic" definition of this notion like model completeness. (A theory is model complete if every formula is equivalent to an existential formula in that theory or, equivalently, if any embedding of its models is elementary.)


Also, I wonder whether near model completeness of a theory implies its inductiveness or not. Recall that a theory is inductive if the union of any chain (or directed family) of its models is again a model. Equivalently, a theory is inductive iff it is $\forall \exists$-axiomatisable. 



Near model completeness
~~I'm still thinking about your main question. Meanwhile here is my answer to your additional question.~~


**A sandwich criterion by Kueker and Turnquist**


You may have seen this before, but it's still worth stating here as it may well be the best answer possible. Proposition 2.3 in [Nearly Model Complete Theories](http://onlinelibrary.wiley.com/doi/10.1002/malq.19990450302/abstract) by Kueker and Turnquist, MLQ 45 (1999), says that near model completeness is equivalent to the following sandwich condition:


Whenever $M\subseteq N\subseteq M'$ are all models of $T$, $\bar a\in M$ and $(M,\bar a)\equiv (M',\bar a)$, then in fact $(M,\bar a)\equiv(N,\bar a)\equiv(M',\bar a)$.


It's formula-free, but it's not a category theoretic definition.


**A nearly model complete theory that is not inductive**


Let $T$ be the theory of infinite discrete linear orders, i.e. of non-empty linear orders in which every element has a successor and a predecessor. Define binary existential formulas $<\_n$ such that $a<\_nb \iff \exists c\_1\ldots c\_n(a < c\_1 < \ldots < c\_n < b)$. $T$ clearly has quantifier elimination down to boolean combinations of these formulas. So $T$ is nearly model complete.


For every $n$, $M\_n=\frac 1{2^n}\mathbb Z\subset\mathbb Q$ is a model of $T$. However, the union of the chain $(M\_n)\_{n<\omega}$ consists of the dyadic rationals and is dense, hence not a model of $T$. So $T$ is not inductive.


