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.)
Best Answer
There are a few good answers to this question, depending on the audience. I've used all of these on occasion.
A way to solve polynomials
We came up with equations like $x - 5 = 0$, what is $x$?, and the naturals solved them (easily). Then we asked, "wait, what about $x + 5 = 0$?" So we invented negative numbers. Then we asked "wait, what about $2x = 1$?" So we invented rational numbers. Then we asked "wait, what about $x^2 = 2$?" so we invented irrational numbers.
Finally, we asked, "wait, what about $x^2 = -1$?" This is the only question that was left, so we decided to invent the "imaginary" numbers to solve it. All the other numbers, at some point, didn't exist and didn't seem "real", but now they're fine. Now that we have imaginary numbers, we can solve every polynomial, so it makes sense that that's the last place to stop.
Pairs of numbers
This explanation goes the route of redefinition. Tell the listener to forget everything he or she knows about imaginary numbers. You're defining a new number system, only now there are always pairs of numbers. Why? For fun. Then go through explaining how addition/multiplication work. Try and find a good "realistic" use of pairs of numbers (many exist).
Then, show that in this system, $(0,1) * (0,1) = (-1,0)$, in other words, we've defined a new system, under which it makes sense to say that $\sqrt{-1} = i$, when $i=(0,1)$. And that's really all there is to imaginary numbers: a definition of a new number system, which makes sense to use in most places. And under that system, there is an answer to $\sqrt{-1}$.
The historical explanation
Explain the history of the imaginary numbers. Showing that mathematicians also fought against them for a long time helps people understand the mathematical process, i.e., that it's all definitions in the end.
I'm a little rusty, but I think there were certain equations that kept having parts of them which used $\sqrt{-1}$, and the mathematicians kept throwing out the equations since there is no such thing.
Then, one mathematician decided to just "roll with it", and kept working, and found out that all those square roots cancelled each other out.
Amazingly, the answer that was left was the correct answer (he was working on finding roots of polynomials, I think). Which lead him to think that there was a valid reason to use $\sqrt{-1}$, even if it took a long time to understand it.