$\because$点$E$为$AC$的中点，

$\therefore S_{\triangle ABE}=\dfrac{1}{2}S_{\triangle ABC}$.

$\because BD:CD=2:3$，

$\therefore S_{\triangle ABD}=\dfrac{2}{5}S_{\triangle ABC}$，

$\because S_{\triangle AOE}-S_{\triangle BOD}=1$，

$\therefore S_{\triangle ABE}=S_{\triangle ABD}=\dfrac{1}{2}S_{\triangle ABC}-\dfrac{2}{5}S_{\triangle ABC}=1$，解得$S_{\triangle ABC}=10$.

故答案为：$10$.

$\because$点$E$为$AC$的中点，

$\therefore S_{\triangle ABE}=\dfrac{1}{2}S_{\triangle ABC}$.

$\because BD:CD=2:3$，

$\therefore S_{\triangle ABD}=\dfrac{2}{5}S_{\triangle ABC}$，

$\because S_{\triangle AOE}-S_{\triangle BOD}=1$，

$\therefore S_{\triangle ABE}=S_{\triangle ABD}=\dfrac{1}{2}S_{\triangle ABC}-\dfrac{2}{5}S_{\triangle ABC}=1$，解得$S_{\triangle ABC}=10$.

故答案为：$10$.