First, I didn't say I could prove the existence of anything but myself!
Second, you don't mean "prove the existence of the uniqueness of addition", you mean simply "prove that the result of addition is unique".
Exactly now you do that depends on what number system you are using.
The simplest is the natural numbers where you take as axioms, the existence of a number "1" and the existence of the "successor" of any number as a one-to-one, onto function from the natural numbers to the natural number without "1" and take the "principle of induction" as an axiom (If a set X contains the number 1 and, whenever it contains the number k, it also contains the successor of k, then X is the set of all natural numbers"
We then define addition by "n+ 1 is the successor of n. If m is not 1, then it is the successor of some number, say p. We define n+ m as "the successor of n+ p".
Now we can prove that m+ n is unique:
Let n be a counting number and let X be the set of all m such that n+ m is unique. n+ 1 is, by definition, the successor of n. That is unique since "successor" is a function. Suppose k is in X. Then n+ successor of k is, by definition,the successor of n+ k. n+ k is unique since k is in X. The successor of n+ k is then unique because "successor" is a function. Therefore the succesor of k is in X. Therefore X is all natural numbers: For every natural number, n+ m is unique for all natural numbers m.
Once we have that we can define the integers by: Let NxN be the set of all pairs of natural numbers. We say that two such pairs, (a,b) and (c,d) are equivalent if and only if a+d= b+ c. It's easy to show that that is an equivalence relation and so partitions the set of pairs into equivalence classes. An "integer" is such an equivalence relation.
We define addition of integers by: If x and y are two integers, that is, two such equivalence classes, choose two "representative" pairs, (a,b) from x, and (c,d) from y. "x+ y" is the equivalence class containing the pair (a+c, b+d). In order to show uniqueness here, we must show that choosing different pairs from x and y would give the same equivalence class for x+ y.
That is, suppose (a,b) and (a',b') are both in equivalence class x and (c,d) and (c',d') are both in equivalence class y. That is, a+b'= b+a' and c+d'= d+c'. Adding those two equations, a+b'+c+d'= b+a'+d+c' so (a+c)+(b'+d')= (b+d)+ (a'+c') showing that (a+c, b+d) is equivalent to (a'+c', b'+ d').
