It is well known the finite commutative rings, $Z/nZ$, are not discretely ordered rings. The axiom $\forall x \forall y \forall z((0<z \land x<y) \rightarrow (x*z < y*z))$ is false for the rings $Z/nZ$. However, these rings do have a total cyclic order.
It is easy to define a partial cyclic order for these rings. A common way of defining a partial cyclic order uses the betweenness operator, $B(x,y,z)$, which means $y$ comes after $x$ and before $z$. A simple way to define a partial cyclic order would be $\forall x(B(x,x+1,x+2))$. Determining whether a partial cyclic order can be extended to a total cyclic order is an NP-complete problem. 3SAT can be converted into a partial cyclic order extension problem.
Assume we start with the standard field axioms and add the following axioms for total cyclic order:
Define Distinct, D(x,y,z), and eliminate troublesome models
D1) $\forall x \forall y \forall z(D(x,y,z) \leftrightarrow (x \neq y \land x \neq z \land y \neq z))$
D2) $\exists x \exists y \exists z(D(x,y,z))$
Partial Cyclic Order, B(x,y,z)
PC1) $\forall x \forall y \forall z(B(x,y,z) \rightarrow B(y,z,x))$
PC2) $\forall x \forall y \forall z(B(x,y,z) \rightarrow \neg B(z,y,x))$
PC3) $\forall w \forall x \forall y \forall z((B(w,x,y) \land B(w,y,z)) \rightarrow B(w,x,z))$
Total Cyclic Order
TC1) $\forall x \forall y \forall z(D(x,y,z) \rightarrow (B(x,y,z) \lor B(z,y,x))$
Link Successor and Addition to Cyclic Order
L1) $\forall x(B(x,x+1,x+2))$
L2) $\forall w \forall x \forall y \forall z(B(x,y,z) \leftrightarrow B(x+w,y+w,z+w))$
The rings $Z/pZ$ where $p$ is prime are models of this theory. By compactness and Lowenhiem-Skolem there exists a countably infinite model of this theory. Are there are any algebraically closed fields that are models of this theory? Are there any fields that do not have a total cyclic order?
Your axioms do not involve multiplication at all (other than referring to $1$, and the only property of $1$ used is that it is a nonzero element), so the existence of a total cyclic order on a field depends only on its additive structure. Since any totally order on a field induces a cyclic order satisfying your axioms and there exist totally ordered fields which have any dimension over $\mathbb{Q}$, it follows that every field of characteristic $0$ can be totally cyclically ordered.
On the other hand, no field of characteristic $2$ can be totally cyclically ordered, since in such a field axiom L1 says $B(x,x+1,x)$, which is incompatible with axiom PC2. I don't know what you can say about fields of characteristic $p>2$.