Two relatively prime integers, $a$ $>$ $2$ and $b$ $>$ $2$, let $(a, b)[{n}]$ denote all primes $p$ which are either $1$ $\mod a$, and or $1$ $\mod b$, and $p$ relatively prime to ab ($n$ denotes the $n$th prime in this set, if defined).
Can it be shown that a product of all consecutive primes in $(a, b)[{n}]$ (to infinity) at some point will be $1$ $\pmod {ab}$?
In other words, would it ever be the case that the product all primes in $(a, b)[{n}]$ to infinity is never $1$ $\pmod {ab}$.
Let $P(a, b)$ be the smallest prime $p$ such that the product of all primes $x$ in $(a, b)[{n}] <= p$ is $1$ $\pmod {ab}$.
For example, $P(3, 4) = 19$, since the product of primes $\le 19$ either congruent $1$ $\pmod 3$ and or $1$ $\pmod 4$ relatively prime to $12$, is $1$ $\pmod {12}$.
The first three smallest pairs, $P(3, 4) = 19$, $P(3, 5) = 103$, $P(3, 7) = 283$.
$(3, 4)$
$5*7*13*17*19$ $=$ 1 $\pmod {12}$
$(3, 5)$
$7*11*13*19*31*37*41*43*61*67*71*73*79*97*101*103$ $=$ $1$ $\pmod {15}$
$(3, 7)$
$13*19*29*31*37*43*61*67*71*73*79*97*103*109*113*127*139*151*157*163*181*193*197*199*211*223*229*239*241*271*277*281*283$ $=$ $1$ $\pmod {21}$
Not an answer, but I programmed a function $q(a,b)$ in PARI/GP and I checked the coprime pairs ($a,b$) with $1\le a<b\le 100$. The largest prime in this range occurs for the pair $(77,97)$.
It is $16403311$.
I will continue the search.
I do not really have an idea how to prove that we always eventually get the desired equivalence, but since $(a,b)$ are coprime the chinese remainder theorem might help.