A Gift Problem for the Year 2018

Consider the multiplicative law on $\Bbb R$ defines by $$x*y =x+y+xy =(x+1)(y+1)-1 $$

you can check that it is associative and commutative on $\Bbb R$. Therefore at the end the remaining number is $$\begin{align}x_0*x_1*x_2*\cdots x_{2018} &= 1*\frac{1}{2}*\frac{1}{3}*\cdots *\frac{1}{2018} \\&=\left[\prod_{i=1}^{2018}(1+x_i)\right]-1\\ &=\left[\prod_{i=1}^{2018}\left(1+\frac{1}{i}\right)\right]-1 \\ &=\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \ldots \cdot \frac{2018+1}{2018}-1=\color{red}{2019-1=2018.} \end{align}$$

Hint: Note that $xy+x+y = (x+1)(y+1) - 1$. In particular, this implies that at any stage of the process, if all the numbers on the board are $\{a_1, \cdots, a_k\}$, then the product $$(a_1 + 1) (a_2 + 1) \cdots (a_k + 1)$$ is invariant, i.e. it doesn't change when you erase two numbers $x$, $y$, and replace them with $xy+x+y$. What does this imply the final number must be?