Does axiomatizability in zeroth-order logic have important consequences? If a theory is equationally axiomatizable, this has important consequences (that are studied e.g. in universal algebra).
However, many theories fail to be equationally axiomatizable - examples include fields, integral domains, and partially ordered sets. Nonetheless, all of these examples are zeroth-order axiomatizable. Does this have important consequences?

For instance, the theory of partially ordered sets is generated by the following axioms.


*

*$x \leq x$

*$(x \leq y) \wedge (y \leq x) \rightarrow x = y$

*$(x \leq y) \wedge (y \leq z) \rightarrow x \leq z$
 A: This is usually called universally axiomatizable because the axioms all have implicit universal quantifiers over the free variables. There are various nice properties of universally axiomatizable theories, which are discussed in introductory model theory books. 
For example, if $M$ and $N$ are structures in a first-order language $L$ and $M$ is a substructure of $N$, then every universal $L(M)$ sentence true in $N$ is true in $M$. This means that if $N$ satisfies a universally axiomatized theory $T$ and $M$ is a substructure of $N$ then $M$ also satisfies $T$.  The Tarski-Łoś theorem of model theory establishes the converse: an arbitrary first-order theory has an axiomatization by universal sentences if and only if the class of models of the theory is closed under taking substructures. 
This immediately implies that e.g. partial orders in the language $(<,=)$ have a universal axiomatization, because a substructure of a partial order in that language is again a partial order in that language. However, dense partial orders do not have a universal axiomatization in that language, because $\mathbb{Z}$ in its usual order is a substructure of $\mathbb{Q}$ in its usual order. 
The theory of fields is not universally axiomatizable in the usual signature of fields $(0,1,+,\cdot, =)$. This is because the axioms stating the existence of inverses are not purely universal (and $\mathbb{Z}$ is a substructure of $\mathbb{R}$ in that language). If one tries to circumvent this by adding unary $-$ and $^{-1}$ functions, that leads to other issues - for example, then $\mathbb{R}$ is no longer a field in the extended language, because one must first specify an arbitrary element for $0^{-1}$, and different choices of $0^{-1}$ lead to non-isomorphic structures all corresponding to $\mathbb{R}$. 
