I. First condition
The question of what integers of form $ax^2+bxy+cy^2$ are closed under multiplication is quite interesting. We give a simple proof that it is the case if $\color{blue}{a=1}$,
$$(p^2 + bpq + cq^2)(r^2 + brs + cs^2) = z_1^2+bz_1z_2+cz_2^2$$
\begin{align} z_1 &= p r - c q s\\ z_2 &= p s + q r + b q s\end{align}
So it remains to consider $a\neq1$. Vladimir Arnold showed that if $ax^2+bxy+cy^2 =1$ for some integer $(x,y)$, then there is multiplicative closure. But it is also true that,
$$(2p^2+pq+3q^2)(2r^2+rs+3s^2)=2x^2+xy+3y^2$$
and $2x^2+xy+3y^2\neq1$ for any integer $(x,y)$. We need a second condition.
II. Second condition
"If there are integers $(r_1, r_2, r_3, r_4)$ such that,
$$ar_1^2+br_1r_2+cr_2^2=ac\\ r_3=(ar_1+br_2)/c\\ r_4=(br_1+cr_2)/a$$
then the quadratic form $ax^2+bxy+cy^2$ has multiplicative closure."
Proof: This is given by the identity,
$$(au^2+buv+cv^2)(ax^2+bxy+cy^2)\left(1+\frac{ar_1^2+br_1r_2+cr_2^2-ac}{ac}\right) = az_1^2+bz_1z_2+cz_2^2$$
where,
\begin{align}z_1 &= u(-r_2x+r_1y)+v(r_1x+r_4y)\\ z_2 &= u(r_3x+r_2y)-v(-r_2x+r_1y)\\ r_3 &= (ar_1+br_2)/c\\ r_4 &= (br_1+cr_2)/a \end{align}
So all we need is to solve $ar_1^2+br_1r_2+cr_2^2-ac=0$ (which also yields integer $r_3,r_4$) and the closure follows.
III. Examples
The quadratic form $2x^2+xy+3y^2$ has discriminant $d=-23$, class number $h(d)=3$,
$$(2u^2+uv+3v^2)(2x^2+xy+3y^2) = 2z_1^2+z_1z_2+3z_2^2$$
\begin{align}z_1 &= u(-x+y)+v(x+2y)\\ z_2 &= u(x+y)-v(-x+y)\end{align}
since $(r_1, r_2, r_3, r_4) = (1,1,1,2)$.
For class number $h(d) = 3, 6, 9$, there are (16, 51, 34) negative fundamental discriminants $d$, respectively. In the data below, for brevity only the first five $d$ for each will be given. The format is $d;\; (a,b,c);\;(r_1, r_2, r_3, r_4)$.
A. Class number 3
\begin{align} -23;&\;(2,1,3);\;(1,1,1,2)\\ -31;&\;(2,1,4);\;(2,0,1,1)\\ -59;&\;(3,1,5);\;(2,-1,1,-1)\\ -83;&\;(3,1,7);\;(2,1,1,3)\\ -107;&\;(3,1,9);\;(3,0,1,1)\\ \end{align}
B. Class number 6
\begin{align} -87;&\;(4,3,6);\;(0,2,1,3)\\ -104;&\;(3,2,9);\;(3,0,1,2)\\ -116;&\;(5,2,6);\;(2,1,2,2)\\ -152;&\;(6,4,7);\;(1,2,2,3)\\ -212;&\;(6,2,9);\;(3,0,2,1)\qquad \end{align}
C. Class number 9
\begin{align} -199;\;&(7, 5, 8);\;(3, -1, 2, 1)\\ -367;\;&(8, 7, 13);\;(-1, 3, 1, 4)\\ -419;\;&(5, 1, 21);\;(4, 1, 1, 5)\\ -491;\;&(5, 3, 25);\;(5, 0, 1, 3)\\ -563;\;&(11, 3, 13);\;(2, -3, 1, -3) \end{align}
and so on.
IV. Question
Q: Is it true that quadratic forms $ax^2+bxy+cy^2$ with $a\neq1$ and negative discriminants with class number $h(d)=3m$, then ALL have multiplicative closure?
Update: As pointed out by Jagy in his answer, I should have focused on quadratic forms in the principal genus. For example, for $d=-23$ with class number 3, the 3 forms are,
$$\color{blue}{(1, 1, 6), (2, -1, 3), (2, 1, 3)}$$
but for $d=-87$ (not prime) with class number 6, the 6 forms are,
$$\color{blue}{(1, 1, 22)}, (2, -1, 11), (2, 1, 11), (3, 3, 8), \color{blue}{(4, -3, 6), (4, 3, 6)}$$
and for $d=-199$ (prime) with class number 9, the 9 forms are,
$$\color{blue}{(1, 1, 50)}, (2, -1, 25), (2, 1, 25), (4, -3, 13), (4, 3, 13), (5, -1, 10), (5, 1, 10), \color{blue}{(7, -5, 8), (7, 5, 8)}$$
and only those in blue have multiplicative closure. Thus, for class number $h(d)=3m>3$, then not all quadratic forms per discriminant $d$ will qualify, but it seems we can find at least three.
