Are $x \cdot 0 = 0$, $x \cdot 1 = x$, and $-(-x) = x$ axioms?

I will assume this is in the context of rings (e.g., real numbers, integers, etc). In this case, the axiom defining $0$ is that $x + 0 = x$ for all $x$. $x*0 = 0$ is a result of this since we have $x*0 = x*(0+0) = x*0 + x*0$ which implies $x*0 = 0$ (canceling one of the $x*0$'s).

I am guessing that for the second one you mean $x*1 = x$. This is a definition (axiom).

The third one is a consequence of the definition of $-x$ being the element such that $x + (-x) = 0$. For then we have $(-x) + x$ is also zero so that $x$ is the negative of $-x$.

The question is more profound than is initially seems, and is really about algebraic structures. The first question you have to ask yourself is where you're working:

In general, addition and multiplication are defined on a structure, which in this case is a set (basically a collection of "things") with two operators we call addition (marked $+$) and multiplication (marked $\cdot$ or $\times$ or $\ast$ or whatever). If this structure holds some properties, which are sometimes called axioms, then it is called a unit ring. The properties are:

1. The set is closed under the operator $+$. That is, if $a$ and $b$ are in $R$, then $a+b$ is also in $R$.
2. The set has a member which we mark as $0$. It has the properties that for every $a$ in $R$, $a+0 = 0$ and $0+a = 0$.
3. The operation $+$ is commutative: $a+b = b+a$.
4. The operation $+$ is associative: $(a+b)+c = a+(b+c)$.
5. Every member has an additive inverse: for every $a$ in $R$ there is some $b$ in $R$ such that $a+b = 0$ (we mark $b$ as $-a$).
6. The set is closed under the operator $*$. That is, if $a$ and $b$ are in $R$, then $a*b$ is also in $R$.
7. The set has a member which we mark as $1$. It has the properties that for every $a$ in $R$, $a*1 = a$ and $1*a = a$.
8. The operation $*$ is associative: $(a*b) * c = a * (b*c)$.
9. Multiplication is distributive over addition: $a * (b+c) = a*b + a*c$ and $(a+b) * c = a*b + a*c$.

While this is a long list, and introduces the operator $+$ which is not even explicitly mentioned in the question, these properties are quite natural. For example, the integers $\{ \ldots, -2, -1, 0, 1, 2, \ldots\}$ we all know and love indeed form a ring. The real numbers also form a ring (in fact they form a field, which means they hold even more properties).

In regard to your question, the identity $x * 1 = x$ (I assume that's what you meant) is in fact an axiom - it is axiom 7. However, the other two identities are results of the other axioms.

First identity: We use axioms 2 and 9 to get $0 * x = (0+0) * x = 0*x + 0*x$ and then by adding $-(0*x)$ (the additive inverse of $0*x$, from axiom 5) to both sides, $0 = 0*x$.

Second identity: As stated in axiom 5, $-(-x)$ is just a notation used which means "the additive inverse of $-x$". To show that $-(-x) = x$ we need to show that $x$ is in fact the additive inverse of $-x$, or in other words that $x + -x = 0$ and $-x + x = 0$. But that's just what axiom 5 says, so we're done.

Last point: You might be wondering why did we have to go and introduce addition to answer a question about multiplication? Well, it so happens that without addition the other two identities are simply not true. For example, if we look at the positive integers $\{1, 2, 3, \ldots\}$ with only multiplication, then there is no $0$ there! Simply put, this is because the positive integers do not form a ring.