$a+b=0$ implies $a=b=0$ Let $N$ be a set of non-negative integers. Of course we know that $a+b=0$ implies that $a=b=0$ for $a, b \in N$.

How do (or can) we prove this fact if we don't know the subtraction or order?

In other words, we can only use the addition and multiplication.

Please give me advise.

EDIT

The addition law mean that for $a, b \in N$, there is an element $a+b$ in $N$ and this operation is associative. 
The multiplication law means that for $a, b \in N$, there is an element $ab$ in $N$ and this operation is associative. 
Also the distribution laws hold.



EDIT2

Let me rephrase the question since I don't want arguments on orders.

Let $N$ be a set with operation $+$ and $\times$.

$N$ is a monoid with the operation $+$ and $\times$ respectively. There is an unit element $0\in N$.

The distribution laws hold as in the case of the set of integers.

Can we prove the fact above with this assumption?
Please set out what you mean by the addition law.  You need the axiom that there is no number whose successor is $0$ or this fails.  That is what distinguishes the integers from the naturals.  It allows you to define order as $x \le y \leftrightarrow \exists (z) x+z=y$
You don't need $+$ or $\times$ as forming monoids on the natural numbers, nor the distributive property.  You can get by with less as follows:

Addition has the property that $(0+b)\leq(a+b)$ and $(a+0)\leq(a+b)$.  This is NOT presupposing the order of the natural numbers, or that no number of the naturals has 0 as a successor, but only a monotonicity property of addition.

We'll also assume the existence of an additive identity.

Consequently, if $(a+b)=0$, then $b=(0+b)\leq(a+b)=0.$  So, $b\leq 0.$  Also, $0=(0+0)\leq (0+b)=b$ by substitution of 0 for a in $(a+0)\leq (a+b)$ and the identity rule.  Thus, $0\leq b.$  So, $0 \leq b\leq 0.$  Consequently, $b=0.$

Similarly, $a=(a+0)\leq(a+b)=0.$  Thus, $a\leq 0$.  Also, $0=(0+0)\leq(a+0)=a.$  So, $0\leq a.$  Thus, $0 \leq a\leq 0.$  Consequently, $a=0.$

Therefore, for an algebraic structure with an identity element "$0$", binary operation "$+$" and where $(0+b)\leq (a+b),$ and $(a+0)\leq (a+b),$ and $a \leq x \leq a \implies x=a,$ it holds that "if $(a+b)=0,$ then $a=b=0$".

As an example of an algebriac structure where this holds, and "$+$" is not natural number addition, let "$+$" denote the maximum of two numbers, and consider $(\{0, 1\}, +)$.  Both suppositions used in the proof above can get verified.