Note that we can factor $n!+(n+1)!$ as $n!\cdot [1+(n+1)] = n!\cdot(n+2)$. Thus, we have \begin{align*} 1!+2!&= 1!\cdot 3 \\ 2!+3!&= 2!\cdot 4 \\ 3!+4!&= 3!\cdot 5 \\ 4!+5!&= 4!\cdot 6 \\ 5!+6!&= 5!\cdot 7 \\ 6!+7!&= 6!\cdot 8 \\ 7!+8!&= 7!\cdot 9 \\ 8!+9!&= 8!\cdot 10 \end{align*} The last two numbers are $ 9\cdot 7!$ and $ (8\cdot 10)\cdot 7!$, so their least common multiple is equal to $ \mathop{\text{lcm}}[9,8\cdot 10]\cdot 7!$. Since $ 9 $ and $ 8\cdot 10 $ are relatively prime, we have $ \mathop{\text{lcm}}[9,8\cdot 10] = 9\cdot 8\cdot 10 $, and so $$\mathop{\text{lcm}}[7!+8!,8!+9!\ ] = 9\cdot 8\cdot 10\cdot 7!= 10!$$. Finally, we note that all the other numbers in our list ($ 1!+2!,2!+3!,\ldots,6!+7!$) are clearly divisors of $ 10!$. So, the least common multiple of all the numbers in our list is $ 10!$. Writing this in the form specified in the problem, we get $ 1\cdot 10!$, so $ a=1 $ and $ b=10 $ and their sum is $ \boxed{11} $.