jmc

Setting $y=1,$ we get $$f(x)=2f(x)-f(x+1)+1,$$so $f(x+1)=f(x)+1$ for all $x\in \mathbb{Q}.$ Then $$\begin{array}{rl}f(x+2)& =f(x+1)+1=f(x)+2,\\ f(x+3)& =f(x+2)+1=f(x)+3,\end{array}$$and so on. In general, $$f(x+n)=f(x)+n$$for all $x\in \mathbb{Q}$ and all integers $n.$

Since $f(1)=2,$ it follows that $$f(n)=n+1$$for all integers $n.$

Let $x=\frac{a}{b},$ where $a$ and $b$ are integers and $b\ne 0.$ Setting $x=\frac{a}{b}$ and $y=b,$ we get $$f(a)=f\left(\frac{a}{b}\right)f(b)-f\left(\frac{a}{b}+b\right)+1.$$Since $f(a)=a+1,$ $f(b)=b+1,$ and $f\left(\frac{a}{b}+b\right)=f\left(\frac{a}{b}\right)+b,$ $$a+1=(b+1)f\left(\frac{a}{b}\right)-f\left(\frac{a}{b}\right)-b+1.$$Solving, we find $$f\left(\frac{a}{b}\right)=\frac{a+b}{b}=\frac{a}{b}+1.$$Therefore, $f(x)=x+1$ for all $x\in \mathbb{Q}.$

We can check that this function works. Therefore, $n=1$ and $s=\frac{3}{2},$ so $n\times s=\boxed{\frac{3}{2}}.$