I have partitioned the prime factors of $p_n\#$ -- using the typical primorial definition: $x\#$ is the product of all primes not greater than $x$ -- into two sets identified by product: $d,\frac{p_n\#}d$; assume that $d\gt 1$ and $\frac{p_n\#}d\gt 1$. I have reduced the problem I am working on down to the following: given $q,n$ such that $(q,p_{n+1}\#)=1$ and taking the sequence of prime factors of $\frac{p_n\#}d$ as $p_{u_i}$,
If there exist non-negative integers $r_{u_i},s_{u_i},i=1,2,\dots,\omega\left(\frac{p_n\#}d\right)$ such that $q\equiv \prod_ip_{u_i}^{r_{u_i}}\mod d$ and $q\equiv \prod_ip_{u_i}^{s_{u_i}}\mod p_{n+1}$, do there exist non-negative integers $t_{u_i}$ such that $q\equiv \prod_ip_{u_i}^{t_{u_i}}\mod dp_{n+1}$?
The potential counter-example I can think of involves $\varphi(d)=\varphi(p_{n+1})$, such as $d=55,p_{n+1}=41$, but where $q$ is a residue in different phases of $d,p_{n+1}$ so that there are no such $t_{u_i}$. Is this a reasonable scenario? If not, is there some way to conclude that $t_{u_i}$ must exist given the above conditions? Note that we can use induction since we have that the integers $t_{u_i}$ exist for $n=3$, so it is possible to work from that base case.
In particular, we can guarantee that $r_{u_i},s_{u_i}$ exist by induction, noting that there is at least one primitive root $z$ modulo $p_{n+1}$ which divides either $d$ or $\frac{p_n\#}d$; swap $d$ with $\frac{p_n\#}d$ if $z\mid d$.
There is a counterexample to this conjecture, therefore as stated it is false. In particular, let $n=4,p_{n+1}=11$ with $d=5\cdot 7$ and $\frac{p_n\#}d=2\cdot 3$. Then some of the $q$ which were constructible under $n=3$ have associated values $q+ap_n\#$ which are not constructible for some $a$ under $n=4$ for this $d$. In particular, note that there are no $a,b\in\Bbb Z^+: 2^a3^b\equiv 211\mod 385$. By contrast, there are $a,b$ such that $5^a7^b\equiv 7\mod 66$ (trivially), but further investigation is required.