In this edit the postcript has been modified.
I will address the the most recent version of the question, which asks about the relationship between the following two features of a logic $L$, but note that a careful discussion of this topic should first clearly define what counts as a logic.
(1) Abstract Completeness of $L$: The set of valid $L$-sentences is recursively/computably enumerable [hereafter: r.e.].
(2) Compactness of $L$: A set $S$ of sentences of $L$ has a model if every finite subset of $S$ has a model.
(1) does not imply (2).
For example, let $Q$ be the quantifier expressing "there are uncountably many", i.e., $Qx\phi(x)$ holds in a structure $\cal{M}$ with universe $M$ iff the set of $m\in M$ such that $\phi(m)$ holds in $\cal{M}$ is uncountable. Let $L_{FO}(Q)$ be the result of augmenting first order logic $L_{FO}$ with the new quantifier $Q$. Vaught proved that the set of valid sentences of $L_{FO}(Q)$ is r.e. Later [1970] in a seminal paper, Keisler gave an elegant axiomatization of $L_{FO}(Q)$.
On the other hand, it is easy to see that $L_{FO}(Q)$ does not satisfy compactness, e.g., for $\alpha < \aleph_1$ introduce constant symbols $c_{\alpha}$ and consider the set $S$ of sentences consisting of $\lnot Qx (x=x)$ [expressing "the universe is not uncountable"] plus sentences of the form $c_{\alpha}\neq c_\beta$ for $\alpha < \beta < \aleph_1$. It is easy to see that every subset of $S$ has a model, but S itself does not have a model.
I should point out that $L_{FO}(Q)$ has a limited form of compactness known as countable compactness: if $S$ is a countable set of sentences of $L_{FO}(Q)$, then S has a model if every finite subset of $S$ has a model [Vaught, ibid].
All of the above features of $L_{FO}(Q)$ [abstractly complete, countably but not fully compact] are shared by a number of other generalized quantifiers, including the stationary quantifier [introduced in the late 1970's and intensely studied in the 1980's]. However, as shown by Shelah, there are other generalized quantifiers that generate fully compact logics that also are abstractly complete
(2) does not imply (1) either.
For example consider the "logic" whose nonlogical symbols are the arithmetical ones, and whose axioms are the usual axioms of first order logic plus all the axioms of true arithmetic , i.e, arithmetical sentences that hold in $\Bbb{N}$. The semantics of the logic is the same as first order logic, so compactness continues to hold; but clearly abstract completeness fails by Tarski's undefinability of truth-theorem, which says that $Th(\Bbb{N})$ is not arithmetical, let alone r.e.
PS A "naturally occurring logic" that also serves to show that (2) does not imply (1) is the existential fragment of second order logic [denoted $ESO$, and which has the same expressive power as "Independence Friendly Logic", see here Section 4.1 of here]. The compactness of $ESO$ follows from the usual proofs of compactness of first order logic [including the ultraproduct proof], but the set of valid sentences $VAL_{ESO}$ of $ESO$ is not even arithmetically definable let alone r.e., since one can "read off" true arithmetic from $VAL_{ESO}$. More specifically, for a sentence $\varphi$ in the language of arithmetic, it is easy to see that $$\mathbb{N}\models \varphi\Leftrightarrow [ \exists X \big( (\mathsf{Q}\land ``~X \mathrm{~is~nonempty~and~has~a~least~element"})\rightarrow \varphi \big)\in VAL_{ESO}].$$ In the above $\mathsf{Q}$ is the conjunction of the finitely many axioms of Robinson arithmetic (Thanks to Noah Schweber for pointing out the argument for nonarithmeticity of $VAL_{ESO}$, and to Rodrigo Freire for catching a blooper; the previous version incorrectly asserted that $VAL_{ESO}$ is co-r.e.).