Pair of prime numbers to have the same product Say we have 2 prime numbers $a$,$b$ and $c=a*b$.
Is there any other pair of prime numbers $x,y$ (distinct from $a$ and $b$) so that $c=x*y$ ? 
 A: Yes, there is, there are infinitely many in fact. Assuming we're talking about $\mathbb Z = \{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots \}$, we have, for example $$14 = 2 \times 7 = -2 \times -7$$ and $$-14 = -2 \times 7 = 2 \times -7.$$
These don't count as distinct factorizations because of multiplication by units. But the way you worded your question, unless you edit it, I think I have given a valid answer.
A: I've read and re-read your question and I see only one indication that you meant to restrict "prime numbers" to the prime numbers among the natural numbers (for this purpose it doesn't matter if you consider 0 a natural number or not).
That would be the fact that you used the tag prime-numbers, which has this official description:

Prime numbers are natural numbers greater than 1 not divisible by any smaller number other than 1. This tag is intended for questions about, related to, or involving prime numbers. 

Let's assume that you did indeed mean such a restriction. Then the answer to your question is no. The natural numbers form a unique factorization domain (UFD) and we don't consider reordering to fundamentally change a factorization. 
That means $ab$ is the same as $ba$, though in general we do prefer $ab$ if $a \leq b$. So for instance $2 \times 3$ and $3 \times 2$ are not distinct factorizations of 6. Well, you did say $x$ and $y$ have to be distinct from $a$ and $b$.
If you allow negative integers, it gets slightly more interesting. Without loss of generality, set $x = -a$ and $y = -b$. Then $ab = xy = c$. That's why it's better to say that ordering and multiplication by units (like $-1$) don't create distinct factorizations.
If you allow $c = 0$, things get a bit screwy. Then, if $a$ and $x$ are both 0, then $b$ and $y$ can be any integers at all as long $b \neq y$.
And if you look into rings of algebraic integers, many of which are not UFDs, things get much, much more interesting. But for the purpose of your question, are we sticking to the weaker definition of primality, which is now more commonly called "irreducible"? If so, the most famous example is $$6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5})$$ from $\mathbb Z[\sqrt{-5}]$. None of those numbers are primes by the stronger definition, but all the numbers to the right of the equals sign are irreducible in $\mathbb Z[\sqrt{-5}]$, they're not divisible by any numbers of smaller norm other than 1 and $-1$.
A: The answer is "sort of".  We can factor $10$ into primes in four ways: $2\times 5$, $5\times 2$, $(-2)\times (-5)$, and $(-5)\times(-2)$.  All of $2,5,-2,-5$ are prime.
However, in the integers, factorization into primes is unique up to (a) order, and (b) multiplication by $-1$.  That is, if we rearrange $5\times 2$, we get $2\times 5$.  If we multiply each of $-2, -5$ by $-1$, we turn $(-2)\times (-5)$ into $2\times 5$.  This is called the Fundamental Theorem of Arithmetic.
If, instead of "integers", we consider different types of numbers, then different properties hold.  The integers are what's called a UFD; in all UFD's a version of the fundamental theorem of arithmetic holds.
A: No.
$c = a*b = x*y$ would mean $a = \frac {x*y}{b}$.  As $b$ is prime that would mean either $b|x$ or $b|y$.  As $x,y$ are prime that would mean $b=x$ or $b = y$.
This is the Unique Factorization Theorem also called the Fundamental Theorem of Algebra.
Notice, however I took for granted, without proof,  that if a prime $b$ divides $x*y$ then $x$ must divide either $x$ or $y$.  This assumption is Euclid's Lemma and is key to proving Unique Factorization Theorem.
