a theory $(x,+,\cdot)$ satisfies $x \cdot x=0$ and $x \cdot (y \cdot z)=(x\cdot y)\cdot z$ I asked the question https://math.stackexchange.com/questions/885917/a-theory-that-satisfies-x-cdot-x-0-and-x-cdot-y-cdot-z-x-cdot-y-cdot-z, but I lost my account since then. So. I am posting my edit as a new question. 
Let there be a first-order theory where there are mathematical objects/sets/numbers in a theory, and the number of mathematical objects in the theory is countably/uncountably infinite. The theory has $+$ and $\cdot$ as language.
First of all, let $x \cdot y$ be represented as $xy$. For every $x$ and $y$, if $x$ and $y$ are different objects in the theory, then $xy \neq 0$ unless $x$ or $y$ is zero. For every $x$, $xx = 0$. 
The following must also all be satisfied:
$0 \cdot 0 = 0$, $0 + 0 = 0$, $(x+y)z = xz+yz$, $x(yz) = (xy)z$, $x+0 = 0+x = x$. $0$ is an object in the theory.
Is there any consistent model of such theory?
At this point, I do get reminded of anticommutative property of exterior algebra. But I am not sure how I would be able to construct a model. 
Edit: What happens if an axiom that $xy = yx$, commutative property is added into the theory?
Edit: I assumed right distributivity only, but it's fine that both left and right distributivity be satisfied, though for the first edit, this would matter.
Edit: $0 + x = x+0 = x$ for every $x$.
Edit: $xy = 0$ if $x=y$ or $x=0$ or $y=0$.
 A: First, let me point out that all the axioms you have written down are equations (Edit: except the one that says $xy = 0\rightarrow x = y \lor x = 0\lor y = 0$... I'll ignore that one for now). That is, they are equations in terms involving variables and constants, which are intended to be universally quantified (i.e. when you write $x + 0 = x$, you mean $\forall x\,x+0 = x$).
A theory in a language consisting of constant and function symbols (no relations) all of whose axioms are equations is called an equational theory, and the class of its models is called a "variety of algebras". These are the objects of study of Universal Algebra. 
Every equational theory is consistent. In fact, every equational theory has a trivial model $\{*\}$ in which every constant is interpreted as $*$ and every function is the constant function returning $*$. All equations are trivially satisfied, since both sides can only evaluate to $*$. The question of whether an equational theory has any nontrivial models can be much harder.
Now let me address some problems with your question:

Let there be a first-order theory where there are mathematical
  objects/sets/numbers in a theory...

This doesn't make any sense. A first-order theory is a collection of axioms - it doesn't make sense to say it has numbers in it. 

...and the number of mathematical objects in the theory is
  countably/uncountably infinite.

I think what you mean here is that a model for the theory should be infinite (also, you don't have to say that the number of mathematical objects in a model is infinite - just say that the model is infinite). The theory you wrote down has finite models, as a pointed out above. If you want to assert that the theory has only infinite models, you should add the following infinite family of axioms to your theory: $$\phi_n: \exists x_1\, \exists x_2\,\dots\,\exists x_n (\bigwedge_{i\neq j} x_i \neq x_j).$$
The sentence $\phi_n$ asserts that there are at least $n$ distinct elements. Taken all together, $\{\phi_n\mid n\in \mathbb{N}\}$ assert that any model is infinite.

The theory has + and ⋅ as language.

This is fine, except that you proceed to use $0$ in your axioms. Any symbol used in the axioms must be a variable, a primitive logical symbol (like $=$), or in the language.

Is there any consistent model of such theory?

It doesn't make sense to ask if a model is consistent. A model exists or it doesn't. If a model exists, then the theory is consistent.
Finally, you've been making a lot of changes to your list of axioms, which suggests that you haven't quite worked out exactly what you want from this theory. Maybe it would help to explain your motivation for considering this theory, and to describe what you want informal terms before jumping right into the list of axioms.
A: EDIT: OP keeps changing the question in response to each answer I post, so I give up.
By your axioms,
$$0=xx=(x+0)x=xx+x0=0+x0=x0$$ and then since you say the only way for $xy$ to equal $0$ is if $x=y$, then it follows that $x=0$. So there is no such algebraic structure other than the one with a single element $0$.
A: A possibility for such axioms (before the edit) are associative Lie algebras, i.e., $2$-step nilpotent Lie algebras. Writing $x\cdot y=[x,y]$ we have $[x,x]=0$ and $[x,y]+[y,x]=0$ (by $[x+y,x+y]=0$).
Furthermore $x\cdot (y\cdot z)=[x,[y,z]]=0=[[x,y],z]=(x\cdot y)\cdot z$, because of $2$-step nilpotent.
