Given curve is $xy+ax+by=0.........\left(1\right)$

It passes through $\left(1,1\right)$

$\therefore a+b=-1........\left(2\right)$

Differentiating $\left(1\right)$ with respect to $x$, we get

$x\frac{dy}{dx}+y+a+b\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}\left(x+b\right)=-\left(y+a\right)$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=\frac{-(1+a)}{1+b}$

Also given that tangent at $\left(1,1\right)$ makes an angle ${\tan }^{-1}2$ with $x$-axis. Let $\theta$ be the angle made by the tangent to the curve at $\left(1,1\right)$.

$\therefore \theta ={\tan }^{-1}\left(2\right)\Rightarrow \tan \theta =\tan \left({\tan }^{-1}2\right)=2$.

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}=2$

$\Rightarrow \frac{-\left(1+a\right)}{1+b}=2$

$\Rightarrow a+2b=-3........\left(3\right)$

By equation $\left(3\right)-$ equation $\left(2\right)$, we get

$b=-2$

Substituting $b=-2$ in $\left(2\right)$, we get

$a=1$

Now,

$\frac{a+b}{ab}=\frac{1+\left(-2\right)}{-2}=\frac{1}{2}$

Hence, option $\left(2\right)$ is correct.