In the book The Moscow Puzzles by Boris A. Kordemsky (ISBN 0-684-14860-6), the puzzle "Different actions, same results" (#52), asks for sets of four and five numbers strictly positive integers which give the same result when added and when multiplied.
With two numbers it's easy to see that the only solution to $a+b=a \cdot b$ is $a=2$, $b=2$, because the relation can be written $b={a \over {a-1}}$ and the right hand side is a strictly decreasing function of $a$.
With three numbers the relation $a+b+c = a \cdot b \cdot c$ has the solution $a=1$, $b=2$, $c=3$, and the wording of the puzzle suggests it is the only solution.
Is there an elementary proof?
We assume without loss of generality that $a \leq b \leq c$.
If $a \geq 2$, then we have $abc \geq bc + bc \geq 2b + c + c > a + b + c$.
Thus $a = 1$ and we want to find $b, c$ such that $1 + b + c = bc$.
This then becomes $(b - 1)(c - 1) = 2$ and the rest is clear.