$\Bbb{S}: \{ ax^2+by^2 \ | \ x, y \in \Bbb{Z} \}$. Find the condition of $a, b$ which makes $\Bbb{S}$ closed for multiplication. 
Let $\Bbb{S}: \{ ax^2+by^2 \ | \ x, y \in \Bbb{Z} \}.$ for the fixed $a, b \in \Bbb{N}$. Then, find the condition of $a, b$ which makes $\Bbb{S}$ closed for multiplication.

\begin{align}
& \text{if } a=b=1: \text{closed for multiplication. (ref)} \\
& \text{if } a=\alpha^2, b=\beta^2: ax^2+by^2=(\alpha x)^2+(\beta y)^2, az^2+bw^2=(\alpha z)^2+(\beta w)^2. \\
\therefore \; & (ax^2+by^2)(az^2+bw^2)=(\alpha^2xz\pm\beta^2yw)^2+(\alpha\beta xw \mp \alpha\beta yz)^2. \\
\Rightarrow \; & \text{closed for multiplication.}
\end{align}
What else can be a condition to $a, b$?
ref) About the set of $\Bbb{S}=\{n|n=a^2+b^2, a, b \in \Bbb{Z}.\}$
Edit) comment told me that the second example doesn't work... Sorry for my mistake.
Comment says:
$a=1: \\ \Bbb{S}: \{ x^2+by^2 | x, y \in \Bbb{Z} \}. (x^2+by^2)(z^2+bw^2)=(xz\pm byw)^2+b(yz\mp xw)^2. \\ \Rightarrow \text{Closed for multiplication.}$
 A: I found a (surprisingly recent) reference that answers this question:

Earnest, A.G. and Fitzgerald, Robert W. "Represented Value Sets for Integral Binary Quadratic Forms and Lattices." (2007).  Proceedings of the American Mathematical Society, 135, 3765–3770.

Corollary 2.5 in particular gives the following necessary and sufficient condition: define $d = \gcd(a,b)$ and let $A = a/d, B = b/d$.  Then $\{ax^2 + by^2\}$ is closed under multiplication iff $d = Ax^2 + By^2$ for some $x,y\in \mathbb Z$.
I said above that the set of $b$ for a given $a$ is infinite iff $a$ is not a square.  This can be seen independently of the above criterion, as I describe below:
It is easy to see by Brahmagupta identity that $k^2 ( x^2 + ny^2 )$ also satisfies the multiplicative property, so if $a=k^2$ then we can take any $b = nk^2$.  On the other hand, if $a$ is not a square, then $a \in \mathcal S$, so in order for $\mathcal S$ to be closed we need $a^2 \in \mathcal S$, hence $a^2 = ax^2 + by^2$ for some $x,y$.  Since $a$ is not a square, we cannot have $y=0$, so then $y^2 \ge 1$ and $b \le a^2$, so there are only finitely many values of $b$.
