Its semigroup.
Now, we check if idenity element exists.
let 'e' be identity element.
$\therefore$ a+ e = ae = a
$\therefore$ a(e-1) = 0
$\therefore$ There is no unique identity when a=0
But, a $\in$ {1, 2, 3......}
Thus, identity element e=1
If inverse were to exist, it would have been multiplicative inverse which is rational number and not natural number.
Hence, inverse doesnt exist.
So, its a monoid and hence, option (B)