The set of the first $n$ consecutive integers has sum $\frac{n(n+1)}{2}$ and product $n!$.
I would like to prove the converse: If a set $S$ of $n$ positive integers has sum $\frac{n(n+1)}{2}$ and product $n!$, then $S$ is the set of the first $n$ consecutive integers.
Note that you only need one of the sum or product to prove what you want. As these are the minimum sum or product of $n$ distinct positive integers, any set of $n$ with this sum or product must be the first $n$. Just sort $S$ and subtract $1,2,3,4,\ldots n$ term by term. If any of these subtractions give a positive value, the sum will be larger than $\frac 12n(n+1)$
This question does not seem to be correct? If all are positive and distinct, it is trivial. If one allows repeated numbers, $1, 3, 3, 4, 4, 4, 7, 10$ is a counterexample for $n=8$. If negative numbers are allowed, $-2, -1, 2, 3, 5, 6, 7, 16$ is another counterexample for $n=8$. I hope that my calculations are not mistaken.
I'd like to add that the argument of minimality is nice but it works because you fixed $n$ expressedly.
If we only search for sets of unspecified number of elements whose sum and product are $T_n$ and $n!$ then $\{1,2,..,n\}$ is not necessarily the only solution, what would happen if we could replace two, three or more elements by a single one ?
The question becomes is it possible to have $a+b=ab$ for any two distinct integers ?
$a+b=ab\iff a=ab-b=a(b-1)$ if $a=0$ then $b=0$ and if $a\neq 0$ then $(b-1)=a/a=1$ so $b=2$ and $a+2=2a\iff a=2$
The only couples are $(0,0)$ and $(2,2)$ but they are not distinct, so we are safe.
But this is not over, what about $a+b+c=abc$ ? $\to$ Find all integral solutions to $a+b+c=abc$.
This time, there is a solution in positive integers which is $(1,2,3)$
So $\{1,2,3,4,5\}$ and $\{4,5,6\}$ would have same sum and product, but of course only the first one has $5$ elements.
And it get worse with more elements, because not only we would have to look at $a+b+c+d=abcd$ but also to grouping of these numbers two by two, and so on.
