Evidently, $\{0\}$ has to be excluded since it has no inverse. My question is reduced to: Is there any total order on complex numbers w/o $0$? From what I sense (but not 100% sure), the lexicographical order (also known as lexical order) is a total order but not linear order.
The answer may be implied by this publication: Levi, F.W. (1942), "Ordered groups.", Proc. Indian Acad. Sci., A16: 256–263.
If this: http://mathworld.wolfram.com/TotallyOrderedSet.html is correct. However, kindly notice that $<$ is a total order for $\mathbb{R} \setminus \{0\}$ and $-1 < 1$ is true but when we multiply both sides by $-1$, we get a rather surprising result $1 < -1$ Hence, it is not a liner order for the multiplication. Any comments? Interpretation of it?
Firstly, the terminology linear order, total order and simple order all have the same meaning. Therefore, linearly ordered group is sometimes also called totally ordered group or simply ordered group. From now on, I will use the terminology linear order and linearly ordered group.
Secondly, as GEdgar has pointed out, the linear order on an abelian linearly ordered group has to be translation-invariant as well. For example, let $(G, \cdot, <)$ be an abelian linearly ordered group, then we must have $a < b$ implying $ac < bc$ for all $a, b, c \in G$.
The answer to your question is no. It is not possible to equip the abelian group $(\mathbb{C} \setminus \{0\}, \cdot)$ with a translation-invariant linear order $<$.
Firstly, we claim that all abelian linearly ordered group $(G, \cdot, <)$ must be torsion-free. (Recall a torsion-free group is a group where all non-identity elements have infinite order.) Suppose not, say $g^n = 1$ for some $g \ne 1, n \in \mathbb{N}$. By trichotomy, either $g < 1$ or $g > 1$. Say $g > 1$. Multiplying $g$ on both sides gives $$g^2 > g$$ $$g^3 > g^2$$ $$\vdots$$ $$1= g^n > g^{n - 1}$$ Chaining these inequalities gives $1 > g^{n - 1} > \cdots > g^2 > g > 1$ which is a contradiction. The proof is similar for $g < 1$.
Finally, observe that $(\mathbb{C} \setminus \{0\}, \cdot)$ is not torsion-free since $-1$ is of order $2$ (see bof's comment). Therefore, it cannot be an abelian linearly ordered group.