I can think of two ways to make this question precise, and they have different answers. I think it is actually interesting and insightful to see these different ways and why their answers may be different. It should also shed light on why the name in the positive setting is justified. So I will work out both.
First an important technique is that of Morleyisation. This is a way of sneaking in some negation in positive formulas. We start with an $h$-inductive theory $T$. For every positive existential formula $\phi(\bar{x})$ we introduce a new relation symbol $N_\phi(\bar{x})$. We then extend $T$ to $T_1$ by adding the $h$-inductive sentences
$$
\forall \bar{x}(\phi(\bar{x}) \wedge N_\phi(\bar{x}) \to \bot)
\quad\text{and}\quad
\forall \bar{x}(\phi(\bar{x}) \vee N_\phi(\bar{x})).
$$
That is, $T_1$ expresses that $N_\phi(\bar{x})$ is equivalent to $\neg \phi(\bar{x})$. We iterate this process to build $T = T_0 \subseteq T_1 \subseteq T_2 \subseteq \ldots$, and we let $T' = \bigcup_{n < \omega} T_n$. Then $T'$ has the property that every positive existential formula has a negation. Let's make up a name for this.
Definition. Let $T$ be an $h$-inductive theory, such that for every positive existential formula $\phi(\bar{x})$ there is a positive existential formula $\psi(\bar{x})$ with
$$T \models \forall \bar{x}(\neg \phi(\bar{x}) \leftrightarrow \psi(\bar{x})).$$
Then we call $T$ fully negated.
Lemma 1. In a fully negated theory, every first-order formula is equivalent to a positive existential formula.
Proof. First we replace all occurrences of $\forall$ and $\to$ by positive connectives, negation and the existential quantifier. Then the argument easily follows by induction on the complexity of the formula, using the fully negated hypothesis for the negation step. QED.
The point of this is that we can express any first-order theory as an $h$-inductive theory, but in a bigger signature. This process is harmless, because the models do not really change. So in this sense positive logic is a strictly more general setting than first-order logic.
There is already a direct link with positive model completeness.
Proposition 2. For a fully negated $h$-inductive theory $T$ every homomorphism is an elementary embedding. Hence every model is positively closed, so $T$ is positively model complete.
Proof. Let $f: M \to N$ be a homomorphism of models of $T$ and let $\phi(\bar{x})$ be a first-order formula. Then by lemma 1 there is positive existential $\psi(\bar{x})$ that is equivalent (modulo $T$) to $\phi(\bar{x})$. So for $\bar{a} \in M$ we have that $M \models \phi(\bar{a})$ iff $M \models \psi(\bar{a})$, which implies $N \models \psi(f(\bar{a}))$ hence $N \models \phi(f(\bar{a}))$. So $f$ is an elementary embedding. QED.
This gives the first interpretation of the question and its answer. If we consider a first-order theory $T$ as an $h$-inductive theory by Morleyising it then we end up with a fully negated theory. So by proposition 2 such a theory is always positively model complete. In other words: if we view positive logic as a generalisation of first-order logic, then this way of defining positive model completeness does not generalise the original notion of model completeness.
The name is still justified, and there is a reasonable scenario where the notions do coincide. First let's prove the converse of Proposition 2.
Proposition 3. Let $T$ be an $h$-inductive theory and let $\phi(\bar{x})$ be a first-order formula, such that for any homomorphism $f: M \to N$ of models of $T$ we have that $M \models \phi(\bar{a})$ implies $N \models \phi(f(\bar{a}))$. Then $\phi(\bar{x})$ is equivalent to a positive existential formula $\psi(\bar{x})$ modulo $T$.
Proof. This is just a very slight generalisation of a classical result and the proof is really the same. See e.g. here (Theorem 5) for a proof. The proof actually goes through word for word, if you replace "diagram" by "positive diagram" and "existential formula" by "positive existential formula". QED.
Corollary 4. The following are equivalent for an $h$-inductive theory $T$:
- $T$ is positively model complete;
- $T$ is fully negated;
- every homomorphism between models of $T$ is an elementary embedding.
Proof. The implication (2) $\implies$ (3) is Proposition 2, and (3) $\implies$ (1) is trivial. The remaining (1) $\implies$ (2) follows from Proposition 3. Truth of negations of positive existential formulas is preserved upwards by immersions. Since every homomorphism is an immersion, we conclude that indeed every negation of a positive existential formula must be equivalent to a positive existential formula. QED.
Now back to the technique of Morleyisation: we can also partially Morleyise a theory, by just adding a negation for each relation symbol. So for every relation symbol $R$ we add $N_R$ and let our $h$-inductive theory express that $N_R$ is equivalent to $\neg R$. This way homomorphisms of models of $T$ are just the usual embeddings. For such a theory, the $h$-inductive sentences are the same as $\forall \exists$-formulas. So we can again view every $\forall \exists$-theory as an $h$-inductive theory.
Corollary 5. A $\forall \exists$-theory $T$, considered as $h$-inductive theory, is positively model complete if and only if it is model complete in the classical sense.
Proof. Homomorphisms are precisely embeddings. So such a theory is model complete in the classical sense if and only if every homomorphism is an elementary embedding, which by corollary 4 is equivalent to being positively model complete. QED.