Answer
See below
Work Step by Step
Let $x=(a,b)$ and $y=(c,d)$ be the vectors in $S$:
$\Rightarrow x+y=(a,b)+(c,d)=(a,a+1)+(b,b+1)=(a+b,a+b+2)$
Hence $x+y \notin S$.
$\Rightarrow S$ is not closed under addition.
Set of Rational numbers is closed under usual addition.
Because our scalars can come from set of real numbers
In particular, let $\lambda$ be a scalar.
Obtain: $\lambda .x=\lambda(a,b) =(\lambda x,\lambda x+\lambda)$
Hence $\lambda x \notin S$
Therefore $S$ is not closed under scalar multiplication.
