$\DeclareMathOperator\Mod{Mod}$Let $S$ be a closed surface and $\Mod(S)$ be its mapping class group.

It is a well known fact, proved in the Primer on Mapping class groups for example, that the subgroup of $\Mod(S)$ generated by two Dehn twists $T_a$ and $T_b$ depends only on the geometric intersection number $i(a,b)$.

Indeed, if $i(a,b) \geq 2$ a ping pong lemma argument on the curve graph shows that $\langle T_a, T_b \rangle$ is a free group. Also, if $i(a,b) = 0$, then $\langle T_a, T_b \rangle$ is clearly free abelian. Lastly, if $i(a,b) = 1$, it can be shown that $\langle T_a, T_b \rangle$ is a braid group on 3 strands, unless $S$ is a torus.

I am wondering whether such a statement can be proven in general, even without knowing a full classification of the groups that can appear. More precisely, if $\{a_1, \cdots, a_n\}$ and $\{b_1, \cdots, b_n\}$ are two collections of curves such that $i(a_i,a_j) = i(b_i,b_j)$ for all $i,j$, is it true that $\langle T_{a_1}, \cdots, T_{a_n} \rangle \simeq \langle T_{b_1}, \cdots, T_{b_n} \rangle$ as subgroups of $\Mod(S)$?.

It seems that the group $\langle T_{a_1}, \cdots, T_{a_n} \rangle$ depends only on a regular neighborhood $N_a$ of the curves $a_i$, which is then homeomorphic to any regular neighborhood $N_b$ of the curves $b_i$. But in general the injection $i \colon N_a \to S$ does not induce an injective map on the mapping class group levels, so this does not seem to give a proper argument.

Any help is greatly appreciated!