Behind these identities is the observation that $x^2+2y^2$ is the norm of the element $z=x+\sqrt{-2}y \in \mathbb{Z}[2i]$ i.e. it is just $(x+\sqrt{-2}y)(x-\sqrt{-2}y)$. So from $N(z_1z_2)=N(z_1)N(z_2)$ it is easy to see that if two numbers $a,b$ are in $S$, then also $ab\in S$. In particular, since clearly $3=1^2+2 \cdot 1^2 \in S$, this shows that $a \in S$ implies $3a \in S$. The same argument still works if we replace $x^2+2y^2$ in the definition of $S$ by any $x^2+ny^2$ ($n$ could also be negative here). But in the problem at hand, we needed to go in the other direction: If $3a \in S$, then $a \in S$. So we are given that $N(z)=3a$ for some $z=x+\sqrt{-2}y \in \mathbb{Z}[\sqrt{-2i}]$ and we somehow would like to divide by $3$, i.e. to consider $N\left(\frac{z}{1+\sqrt{-2}}\right)=3$. This is true, but the main point to check is whether $\frac{z}{1+\sqrt{-2}} \in \mathbb{Z}[\sqrt{-2}]$ where a priori there could be some fractions with $3$ in the denominator (exactly those in my solution above). So here our argument above is simple, but the reason why it really works is that $\mathbb{Z}[\sqrt{-2}]$ is a Unique Factorization Domain. One of the big moments in the beginning history of algebraic number theory was the discovery that this is not true anymore in more general $\mathbb{Z}[\sqrt{-n}]$. So here we really used the special structure of $x^2+2y^2$...!