The order of the ring's product

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$ $\DeclareMathOperator{\GF}{GF}$With your definition for the product of subrings (I take it that in your formula $$A_1 \cdot A_2 = \Set{\sum a_{1i}a_{2j} \mid a_{1i} \in A_1 ,a_{2j} \in A_2 }$$ $A_{1}, A_{2}$ are subrings of a common ring $A$), consider the following example.

Let $A = \GF(p^{6}),$ where $p$ is a prime, $A_{1} = \GF(p^{2}), A_{2} = \GF(p^{3})$, so that $A_{1} \cap A_{2} = \GF(p)$. As to $A_{1} \cdot A_{2}$, it is a subring of $A$ containing $A_{1}, A_{2}$, and thus $A_{1} \cdot A_{2} = A$. But $$p^{6} = \Size{A} \ne \frac{\Size{A_{1}} \cdot \Size{A_{2}}}{\Size{A_{1} \cap A_{2}}} = \frac{p^{2} \cdot p^{3}}{p} = p^{4}.$$

Does this count? I’m not sure since these two rings don’t share identity with the containing ring. But if admissible, it has the advantage of simplicity.

You can take the field $F_2$ of two elements, and the rings $A_1=F_2\times \{0\}$ and $A_2=\{0\}\times F_2$.

Then $|A_1\cdot A_2|=1$, but $|A_1||A_2|/|A_1\cap A_2|=4$.