Added: Unfortunately, the answer below as originally written was wrong. It answered a related, but easier, question. I have added some bracketed remarks to point out where the blunder happens, and appended some remarks about the expected answer to the actual question.
It might help to explicitly reformulate the question. (This is implicit in the proof that the question is equivalent to the twin prime conjecture for $a = 6$, but it seems worthwhile to make it explicit.)
If $k = a m n \pm m \pm n,$
$a k \pm 1 = (a m \pm 1) (a n \pm 1).$
If $a = 2$ then we can write $1 = (2\cdot 1 - 1)$, and so for any $k$,
we have $2 k - 1 = (2 \cdot 1 - 1)(2 k -1)$,
and (as the OP noted) every integer $k$ can be written in the desired form.
Thus we suppose from now on that $a > 2$.
The problem is then equivalent to asking if we can find infinitely many
numbers congruent to $\pm 1 \bmod a$ which have no proper factors congruent
to $\pm 1 \bmod a$. [Added: This is wrong. The problem is rather to
find infintely many multiples $a k$ of $a$ so that both $a k +1 $ and
$a k - 1$ have no proper factors congruent to $\pm 1 \bmod a$.]
This shows that the answer to the question depends a lot on whether or not
every residue class mod $a$ which is coprime to $a$ is congruent to $\pm 1 \bmod a$, i.e. on whether or not $\varphi(a) = 2$. [Added: It is not so clear to me now what difference the condition on $\varphi(a)$ makes, if any.]
If $\varphi(a) = 2$, then we are simply asking whether we can find infinitely many numbers coprime to $a$, which differ by $2$, and which have no proper factors, i.e. we asking for infinitely many twin primes. This is a well-known open problem!
On the other hand,
suppose that $\varphi(a) > 2$ (which in particular is the case if $a > 6$). We may then choose
$b$ and $c$ so that $b, c \not\equiv \pm 1 \bmod a$, and $b c \equiv 1 \bmod a$.
By Dirichlet's theorem, we may choose infinitely primes $p$ and $q$ such that
$p \equiv b \bmod a$ and $q \equiv c \bmod a$.
For any such $p$ and $q$, the product $pq \equiv 1 \bmod a$, but by construction has no proper factors congruent to $\pm 1 \bmod a$.
So if we set $k = (p q - 1)/a$, then $k$ cannot be written in the given form.
This gives infinitely many $k$, as desired.
[Added: Rather, this gives infinitely many $k$ that are not of the form
$a m n + m + n$ or $a m n - m - n$, but doesn't address whether these can be written in the form $a m n - m + n$, since we don't have any control over $a k - 1 = p q - 2.]
It is not clear to me how to gain control over the factors of $p q - 2$, other than to assume it is prime.
So one way to solve the problem would be to find infinitely many primes $r \equiv -1 \bmod a,$ such that $r + 2$ is either prime, or is of the form
$r + 2 = p q,$ where $p,q \not\equiv \pm 1 \bmod a$.
This is then related to Chen's theorem, indeed is a strengthening of it. Thus
we are led to MO Scribe's question on MO which was inspired by the initial posting of the present question on MO.
There is no doubt that this result (on the existence of such primes $r$) is true. (Indeed, standard conjectures predict that there should be infinitely many twin primes $r, r+2$ with $r\equiv -1 \bmod a$.) But it is not clear (to me) whether or not it is in reach of current methods; that is the thrust of MO Scribe's question.
So what is the conclusion: Well, there is no doubt that one should be able to find infintely many such $k$, since it follows from standard conjectures on twin primes satisfying congruence conditions. On the other hand, proving this may be tricky, since it seems to require results that are at the edge of what is currently possible via seiving techniques.