The current question asks for the existence of $L \in \mathbb{N}$ that satisfies the conditions of this question and an additional condition;
I hope that it is still possible to find $L \in \mathbb{N}$ (= hopefully, the additional condition will not cause too much trouble).
Let $u,v,m,n,\epsilon \in \mathbb{N}-\{0\}$ satisfy:
$m < n$; $\gcd(u,m)=1$; $\gcd(v,n)=1$.
Is it possible to find $L \in \mathbb{N}$, such that the following two conditions are satisfied:
(i) $\gcd(u+Lm,v+Ln)=1$.
(ii) $\epsilon \nmid u+Lm$ and $\epsilon \nmid v+Ln$.
A special case: I do not mind to first concentrate on the special case where $\epsilon$ is a prime number (though I slightly prefer $\epsilon$ to be any natural number).
Almost an answer: According to the above mentioned question, it is possible to find $L \in \mathbb{N}$ such that $u+Lm < v+Ln$ and $v+Ln$ is a prime number,
so (i) is satisfied (= the gcd of a prime number and a smaller number is $1$), but it seems that (ii) may not be satisfied: Although we can take $L$ large enough such that (in addition to $v+Ln$ being prime, we also have) $\epsilon < v+Ln$, so $\epsilon \nmid v+Ln$ (because $v+Ln$ is prime), but unfortunately it may happen that $\epsilon | u+Lm$.
Notice that making the smaller of the two, $u+Lm$, a prime $p$, may be worse, because in that case even condition (i) may not be satisfied, since now $\gcd(u+Lm,v+Ln) \in \{1,p\}$.
A remark: Please see this question.
Edit: In view of this question, I think that I will further assume that $\epsilon \geq 3$.
Edit 2: Please see this question; if we wish to apply its answer to my current question, then (if I am not wrong) we will obtain that a positive answer to my current question may depend on $v,n$ and the set of natural numbers which make $v+Ln$ a prime number. (But perhaps a different approach may yield a positive answer to my current question?).
Thank you very much!
Best Answer
As noted in the OP edit, we need to take $\epsilon > 1$ odd as at least one of $1+L$, $2+3L$ is always even.
Suppose $p$ is a prime divisor of $(u+Lm, v+Ln)$. Then it also divides the integer combination $n(u+Lm)-m(v+Ln) = nu - mv$, which is a constant independent of $L$.
So the only possible primes dividing the gcd are the divisors of $nu-mv$. Note that the constraints preclude this quantity from being zero (otherwise $n/v = m/u$ with both fractions being in lowest terms, which contradicts $m\ne n$), so there are only finitely many such primes. Likewise there are only finitely many primes dividing $\epsilon$. Let $P$ be the largest of all primes in either category.
By the argument of the related question, for any prime $p\ge 3$, we can choose a congruence class for $L$ mod $p$ such that $p \nmid m+Lu$ and $p \nmid n + Lv$. For $p=2$, we can’t guarantee the same, but we can at least ensure one of $m+Lu$ and $n+Lv$ is odd by fixing the parity of $L$.
Combining all these congruences for all primes $p\le P$, we get a congruence class for $L$ (modulo the primorial of $P$) which ensures $(m+Lu, n+Lv) = 1$ (recall that primes above $P$ can’t divide the gcd), as well as the $\epsilon$ condition.
I feel there is probably a more efficient construction based on a similar question I have seen years before. Maybe someone can provide it.