There is a list of $n$ numbers. We pick any two numbers, $u$ and $v$ and replace them by $uv + u + v$. Does the final answer after $n-1$ operations, depend on the initial choice.
I noticed that if each $u$ and $v$ is replaced by its product,at the end of $n-1$ operations, we have $a_1\cdot a_2 \cdot \dots a_n$.
Similarly, if we only replace each two numbers by their sum, we get $a_1 + a_2 + \dots + a_n$
But, I am not able to wrap these two facts into an elegant argument and prove that the answer to this question is no.
It only depends on the associativity of the operation. For example, suppose you have three numbers $a_1,a_2,a_3$. Then note that suppose you apply the operation first on $a_2$ and $a_3$, and then $a_1$ and the result, then we write the final result as $(a_1 \cdot (a_2 \cdot a_3))$, while suppose you apply the operation first on $a_1$ and $a_2$, and then $a_3$ and the result, then we write the final result as $((a_1 \cdot a_2) \cdot a_3)$.We need to check whether $(a_1 \cdot (a_2 \cdot a_3))=((a_1 \cdot a_2) \cdot a_3)$. This property is called associativity, and you can check that addition and multiplication satisfy this property.
Now, it doesn't matter how many $a_n$ there are: if your operation is associative, you can prove that the answer at the end will be the same regardless of the order of operations using induction on $n$. I've done it for just $3$.
Call our operation $(u,v) \to (u + v + uv)$ as $\cdot$, and we will check whether $\cdot$ is associative or not. $$ (a_1 \cdot a_2) \cdot a_3 = (a_1 + a_2 + a_1a_2) \cdot a_3 = a_1+a_2+a_1a_2+a_3 + a_1a_3+a_2a_3+a_1a_2a_3. $$ For the other, $$ a_1 \cdot (a_2 \cdot a_3) = a_1 \cdot (a_2+a_3+a_2a_3) = a_1+a_2+a_3+a_2a_3 + a_1a_2+a_1a_3+a_1a_2a_3. $$
It follows that the right hand side of both is the same, so your operation is associative and the answer at the end of your operation will be independent of your order of application.
This will not apply for something like subtraction or division,for example, because subtraction or division is not associative.
Just as an example, $(1 \cdot 2) \cdot 3 = (1+2+2)\cdot 3 = 5 + 3 + 15 = 23$ and $1 \cdot (2 \cdot 3) = 1 \cdot (2 + 3 +6) = 1 \cdot (11) = 1 + 11 + 11=23$.