Question: Simple example in order to clarify what I mean:

Sequence - $\{2 , -1, 4, 2, 9 \}$
Sum and product combination: $\{ 16, -144 \}$

Question:
Is a combination of sum and product unique for any\* sequence?

* "any sequence" except that one which consists only of zeros.

**UPD:** The answer to my initial question is "no". Thanks.
But what's the answer if we restrict ourselves to sequences of the same length and made of non-negative integers? Do two or more non-negative integer sequences exist, each of which has the same {sum,product} combination and the same length?

**UPD:** Both answers are "no". Thanks a lot.

Answer: 
Not in general. Knowing the sum and product of a sequence $r\_1, ... r\_n$ only fixes two coefficients of the polynomial $(x - r\_1)...(x - r\_n) = x^n + a\_{n-1} x^{n-1} + ... + a\_0$, namely $a\_0$ and $a\_{n-1}$, and for $n > 2$ one can choose the other coefficients arbitrarily and solve for the complex roots of the resulting polynomial.

With certain restrictions, this might sometimes be possible, for example if you restrict the sequence to be integers.
Let's say we have a finite tuple (not set) of positive integers, $A = (a\_1,a\_2,\dotsb,a\_n)$, with $\sum a\_i = s$, $\prod a\_i = p$. Consider the simplest prime factorization $p = p\_1^{n\_1} \dotsb p\_k^{n\_k}$ and construct $B = (1,\dotsb,1,p\_1,\dotsb,p\_1,p\_2,\dotsb,p\_k,\dotsb,p\_k)$ where each $p\_i$ occurs $n\_i$ times and we fill out with enough $1$s to make $\sum B = s$ (why can we do this?).

Proving that this produces a $B$ with different elements than $A$ in "most" cases is left as an exercise for the reader. As is extending it to the case where $A$ may contain negative integers. In that case you want to fill up with a clever combination of $1$s and $-1$s.