Since we start with an axiom C: $(\exists i)i^2=-1$ which does not explicitly contradict any other axiom in $\mathbb R$ (since $i$ does not appear in other axioms), there is no way to directly disprove it. But in a slight modification to the aforementioned, there are some axioms which are actually disproven by (C), like trichotomy:
$$(\forall x)[x<0\vee x=0\vee x>0]$$
(read "every number is either negative, positive, or zero".) Because $i^2=-1$ and $0^2=0$ implies $i\neq0$, $i>0$ implies $i^2=-1>0$ implies $1<0$ implies $1^2>0$ which is a contradiction, and $i<0$ implies $i^2=-1>0$ similarly. What we do, then, is we just drop the offending axioms. Importantly, $\mathbb C$ does not contradict the "important" axioms, like commutativity and associativity of addition and multiplication, and the existence of inverses to non-zero elements. The important part is that although we can't keep everything, we can keep some things, and we can work usefully with what remains.
The problem with your example is that you have not just defined a new element $x$, but you have also defined how it multiplies, and your definition contradicts another one which has already been defined. In $\mathbb C$, we just define the part that can not otherwise be derived, and let the other axioms "figure out the rest", so that we don't have any danger of redefining things incorrectly.
By the way, you asked on the previous question what the axioms of the reals are, so I thought I'd list them here, courtesy of Spivak's Calculus.
- (addition is associative) $(\forall a,b,c)\ a+(b+c)=(a+b)+c$
- (additive identity) $(\exists0)(\forall a)\ a+0=a$
- (additive inverse) $(\forall a)(\exists b)\ a+b=0$
- (addition is commutative) $(\forall a,b)\ a+b=b+a$
- (multiplication is associative) $(\forall a,b,c)\ a(bc)=(ab)c$
- (multiplicative identity) $(\exists1)(\forall a)\ a\cdot1=a$
- (multiplicative inverse) $(\forall a\ne0)(\exists b)\ ab=1$
- (multiplication is commutative) $(\forall a,b)\ ab=ba$
- (distributive law) $(\forall a,b,c)\ a(b+c)=ab+ac$
- (trichotomy) $(\forall x)[x<0\vee x=0\vee x>0]$
- (positives closed under addition) $(\forall a>0,b>0)[a+b>0]$
- (positives closed under multiplication) $(\forall a>0,b>0)[ab>0]$
- (least upper bound) $(\forall S\subseteq\mathbb R)[\mbox{if }S\mbox{ has an upper bound, then }S\mbox{ has a least upper bound}]$
The first four just define all the "nice" properties of addition, the next four do the same for multiplication, number 9 covers the distributive property, 10-12 cover the semantics of an order relation, and number 13 "fills in the gaps" in the rational numbers. That last one is a bit complicated to write in symbols, but it allows you to say that all sorts of numbers like $\sqrt2$ and $\pi$ are actually numbers. The relationship with $\mathbb C$ is that numbers 10-12 are thrown away, but the first 9, which define a field, are still okay. Number 13 makes explicit reference to $\mathbb R$, and it needs an ordering to work properly, so we just leave that one as-is. That way you can still have numbers like $i\pi$ and identify them as elements of $\mathbb C$.
I know you didn't want so much notation all at once, but nothing in here is too out there to get from Intro Calc. (By the way, $\forall$ means "for all", $\exists$ means "there exists", and $\vee$ means "or", in case you haven't seen those before.)
If we had an order on the complex numbers, then either $i \prec 0$ or $0 \prec i$.
If $0 \prec i$, then $$0i \prec ii \implies 0 \prec -1$$
Then since $0 \prec -1$, we see that $0 \prec (-1)^2 = 1$. Using (iii) we get
$$0 \prec -1 \implies 1 = 0 + 1 \prec -1 + 1 = 0 \implies 1 \prec 0 \prec 1$$
contradicts (i). The case that $i \prec 0$ is similar. Just use (ii) and add $(-i)$ both sides.
Best Answer
Axioms 6, 7 and 8 are not sufficient for excluding the possibility to order the complex numbers.
If you define $a+bi\prec c+di$ ($a,b,c,d\in\mathbb{R}$) when either $a<c$ or $a=c$ and $b<d$, you get an order relation satisfying those axioms.
The contradiction will show up only if you add another axiom:
First step: proving that $0<1$.
There are two cases: $1<0$ or $0<1$. Suppose $1<0$; then $1-1<0-1$, so $0<-1$. Hence $0(-1)<(-1)(-1)$, that is $0<1$: a contradiction.
Second step: proving that $-1<0$
Since $0<1$, we have $0-1<1-1$.
Third step: getting a contradiction
Suppose $0<i$; then $0i<i^2$, that is, $0<-1$, a contradiction.
Suppose $i<0$; then $i-i<0-i$ and $0<-i$; then $0(-i)<(-i)^2$, that is, $0<-1$, a contradiction.
Conclusion
Axiom 6 cannot hold for $x=0$ and $y=i$.