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(Caveat: normally I wouldn't answer a question with such a limited knowledge of the general theory, but classical analytic number theory seems not so well represented by active MO members.)

 Suppose that $A$ is a finite abelian group. Then I claim that given any set of at least $|A| + 1$ (not necessarily distinct) elements of $A$ one can find a proper subset whose sum is the identity. Proof: Denote the elements $a_i$ for $i = 1$ to $|A|+1$. By the pigeonhole principle, either one of the $|A|$ sums $\sum_{i=1}^{r} a_i$ for $r = 1$ to $|A|$ is the identity, or two of the sums are the same element of $|A|$, in which case  consider the difference. We deduce from this the following: Let $n$ be any integer coprime to $a$ with more than $r:=|(\Z/a \Z)^{\times}|$ prime factors. Then $n$ has a proper divisor of the form $1 \mod a$. 

Suppose that $k$ cannot be represented by the form $amn \pm m \pm n$, and suppose that $a > 2$. It is simple to deduce that this is equivalent to asking that $ak+1$ and $ak-1$ have no proper divisors of the form $\pm 1 \mod a$. It follows that $ak+1$ and $ak-1$ each have _at most_ $r=|(\Z/a \Z)^{\times}|$ prime 
factors. The integers with at most $r$ prime factors
are sometimes called _$r$-almost primes_. If $\pi_r(x)$ counts the number of $r$-almost
primes $\le x$ then 
$$\pi_r(x)  \sim \frac{x (\log \log x)^{r-1}}{\log(x)}.$$
(Compare this to the prime number theorem when $r = 1$.) In particular, we see that the
 $r$-almost primes have zero density (in any sense), and thus:

>2) The density of integers that can not be represented in the form $amn + m + n$ is zero.
Similarly, the density of integers that cannot be represented in the form $amn + m -n$ is zero. In particular, the density of the _$a$-asterios_ numbers, the integers that can neither
be represented in the form $amn+m+n$ _nor_ $amn+m-n$, is zero.

Let $\pi_{r,2}(x)$ denote the number of _twin_ $r$-almost primes less than or equal to $x$, that is, the number of integers $n \le x$ such that $n$ and $n+2$
are _both_ $r$-almost primes. 
(For example, $\pi_{1,2}(x)$ counts the number of twin primes less than $x$.)
What do we know about this function? Brun was the first to give an upper bound
for $\pi_{r,2}(x)$ using sieving techniques. Refinements by others (in particular
Selberg) allowed one to obtain the estimate
$$\pi_{1,2}(x) \ll \frac{x}{(\log x)^2},$$
which gives the correct (conjectural) order of magnitude.
Without going into the Selberg sieve, let me say that what these arguments 
really
give is decent upper _and_ lower bounds of the following kind (for large $x$):
$$\frac{A x}{(\log x)^2} <
\left\{n < x, \ p \nmid n(n+2) \ \text{if} \ p < x^{\alpha}\right\}
< \frac{B x}{(\log x)^2}$$
for non-zero constants $A$ and $B$,
where $0 < \alpha < 1$ is some fixed small constant, which we might imagine for the
sake of argument is $1/10$.  Since every twin prime $>x^{\alpha}$ contributes to
this sum, this gives the correct (up to a constant) upper bound for
$\pi_{1,2}(x)$. It also gives a _lower_ bound for $\pi_{10,2}(x)$, since if every
factor of $n < x$ is at least $x^{1/10}$, then $n$ has at most $10$ prime factors.

The motto I learnt about sieving was the following: upper bounds are easy, lower bounds are hard. Thus, since we are interested in bounding $\pi_{r,2}(x)$, it seems that we are in good shape.
However, there is a subtlety here. Let $\pi(x,z)$ denote the number of integers $n$ less
than $x$ such that every prime factor of $n$ is at least $z$. It's clear by the
argument of the last paragraph that
$\pi_r(x) \ge \pi(x,x^{1/r})$. One might imagine that these numbers are roughly of the
same magnitute. However, it turns out that $\pi_r(x)$ is much bigger than $\pi(x,x^{1/r})$.
The latter is comparable to the number of primes less than $x$, where the former has
an extra factor of $(\log \log x)^{r-1}$. The reason is that $\pi_r(x)$ is dominated
by numbers with (a few) small prime factors. In fact, 
as Kowalski pointed out to me, it is not even obvious that
one can easily obtain the
correct upper bound for $\pi_r(x)$ simply by sieving over primes.
From the asymptotic for $\pi_{r}(x)$, one expects that
$$(*): \qquad \pi_{r,2}(x) =^{?} \  O\left(\frac{x (\log \log x)^{2r-2}}{(\log x)^2}\right).$$ 
(I expect that such a result,  perhaps slightly weaker, _does_ exist
in the literature, I will update this answer when I find the relevant
links - actual experts are being consulted as we speak.)
All one needs to answer 3) is that 
 the exponent of $\log(x)$ in the denominator is $> 1$. 

>3). Assuming the expected result (*), the inverse sum of the _$a$-asterios_ primes converges.

Consider the set of integers $S_a$
which do not have any prime factors of the form
$\pm 1 \mod a$. This is a reasonable thing to do whenever
$|(\Z/a\Z)^{\times}| > 2$. This is a weaker condition, so there are more of these numbers 
and consequently obtaining upper bounds is harder. 
We may form the Dirichlet series
$$L(s) = \sum_{S_a} \frac{1}{n^s}$$
which has an Euler product:
$$L(s) = \prod_{p \not\equiv \pm 1} \left(1 - \frac{1}{p^s}\right)^{-1}$$
Now let $K = \Q(\zeta_a)^{+}$ be the totally real subfield of $\Q(\zeta_a)$.
It has degree $r/2 = \phi(a)/2$, where $r > 2$ unless $a = 1,2,3,4$ or $6$.
(What we say now only makes sense for $r \ge 2$, in which case $r/2 \in \Z$.)
A prime splits completely in $K$ if and only if it is of the form
$\pm 1 \mod a$. Looking at the Euler product of $\zeta_K(s)$, we see that,
up to a constant 
 which can be explicitly written as some product over primes,
$$\zeta_K(s) L(s)^{r/2} \sim  \zeta_{\Q}(s)^{r/2}$$
as $s \rightarrow 1^{+}$, and hence $L(s) \sim (s-1)^{(2-r)/r}$ 
(up to some constant)
 as $s \rightarrow 1^{+}$. We deduce (Perron's formula) that the number
of integers $\le x$
all of whose prime factors are not of the form $\pm 1 \mod a$ is
asymptotic to
$$ \kappa  \cdot \frac{x}{(\log x)^{2/r}},$$
for some non-zero
constant $\kappa$. This is the same analysis that gives the asymptotic
formula for the number of integers $\le x$ which can be written as a sum
of two squares (a result of Landau). 
We immediately deduce:

>4a) The number of integers $a$ such that $ak+1$ (or $ak-1$) has no
prime factors of the form $\pm 1 \mod a$ has zero density.

If $r > 2$ (so $a \ge 3$ and $a \ne 3,4,6$) then the power $2/r$ of
 $(\log x)$ is at most $1/2$. Thus,  we actually
are led to the following guess:

>4b) If $r = \phi(a) > 2$, then one would heuristically expect the inverse
sum of integers $k$ such that none of the prime factors of $ak-1$ and $ak+1$ 
are $\pm 1 \mod a$ diverges. If $r = 2$, so $a = 3$, $4$, or $6$, then (Brun) the series converges.

>Here is a related problem of the very same kind: can one count the number of
integers $n \le x$ such that both $n$ and $n+1$ can be expressed as the sum of
two squares, and prove that there are $\sim x/\log(x)$ such integers
(perhaps up to non-zero constant factors)?


Any integer $n$ can be written as the product of two numbers not
of the form $\pm 1$ _unless_ every prime factor of $n$ is of the
form $\pm 1 \mod a$. The integers all of whose prime factors are
$\pm 1 \mod a$ can be analyzed exactly as in the last paragraph.
In this case, the number of integers all of whose prime factors are of the
form $\pm 1 \mod a$ is asymptotic to
$$ \kappa \cdot \frac{x}{(\log x)^{(1-2/r)}}.$$
Suppose that $r > 2$. Then the set of such integers has density zero, and thus the
set of integers which have a factor _not_ of the form $\pm 1 \mod a$
has density one. Any set of density one has infinitely many 
"twins" satisfying any fixed congruence condition.
Hence:

>5) If $r = \phi(a) > 2$, then there are infinitely  many $k$ such that both
$ak+1$ and $ak-1$  have a factor _not_ of the form $\pm 1 \mod a$. Indeed, such numbers have density one.

Finally, I have nothing to say about problem 1) besides the remarks I made in my rephrasing of the original question here:

http://mathoverflow.net/questions/49751/chens-theorem-with-congruence-conditions