The answer is yes. Let $p_{ij}$, $0\le i,j\le 2023$, be a sequence of distinct primes all of whom are sufficiently large (in particular, larger than 2023). Choosing $a$. Set $a$ such that $a\equiv p_{ij}-i\pmod{p_{ij}^2}$ for any $i,j$. Such an $a$ exists due to CRT. Moreover, due to size constraints, $p_{ij}$ divides exactly one $a_i$, where $a_i = a+i$. Now let \[ K = \left(\prod_{0\le i\le 2023}a_i\right)/\left(\prod_{ij}p_{ij}\right). \]Clearly, $K\in\mathbb{N}$ and moreover, $(K,p_{ij})=1$ for any $i,j$. Choosing $b$. Now let $b$ such that $b\equiv 0\pmod{K}$ and $b\equiv -j\pmod{p_{ij}}$. Once again, such a $b$ exists due to CRT. Moreover, under this, it is clear $\textstyle \prod_i a_i\mid \prod_j b_j$, where $b_j=b+j$. Lastly, if $i\ne i'$ and $j\ne j'$, then we must have $a_i\nmid a_{i'}$, $a_i\nmid b_j$ and $b_j\nmid b_{j'}$ due to the fact each number contains a prime divisor not dividing the other (easily seen). This completes the proof.