Here is a partial answer to your question. Basically, the methods of Goldston, Graham, Pintz, and Yildirim (see here) have some bearing on your question.
Look at the series
$$\sum_{N < n < 2N} \bigg( \chi_1(n) + \chi_1(n + 2) + \chi_2(n + 2) \bigg)
\bigg( \sum_d \lambda_d \bigg)^2$$
where the $\lambda_d$ are the usual Selberg sieve coefficients or some variant, and $\chi_1$ and $\chi_2$ are the characteristic functions of products of exactly one or two primes. Then I worked out a back of a napkin calculation using Theorems 7-9 in GGPY (possibly I made some mistake), and on the Elliott-Halberstam conjecture (a big assumption!) the above is positive, which gives another proof of Chen's theorem (this one conditional).
One nice feature of the GGPY sieve is that it is compatible with the kind of conditions you describe. For example, let $\chi_2$ be the characteristic function of $E_2$ numbers whose prime factors are both 1 mod 4, or split completely in your favorite Galois extension, or some other appropriate condition. (See the GGPY paper as well as a followup which I wrote.) Then the machinery still works. You just multiply the contribution from the $E_2$ numbers by a fourth. In fact you can do better and only sieve on integers $n$ congruent to 1 mod 4, in which case you only have to divide by a half.
According to my napkin calculation, this is numerically not enough, the series is asymptotically slightly negative, and so this argument fails to prove your assertion, even on EH. So perhaps your question really is "difficult" in the way suggested by the parity problem. But this at least shows you how you can hope to prove statements related to what you asked for.
I will try to convince you that no Euclidean proof can possibly show that there are infinitely many primes which are $2 \mod 5$. After giving some definitions, I will explain what I will actually show:
Let $G$ be a finite group. We define an equivalence relation on $G$ by $g \sim h$ if the (cyclic) subgroups generated by $g$ and by $h$ are conjugate. The equivalence classes for this relation are called divisions. Observe that a map of groups always respects this equivalence relation. A more precise (but not yet precise) claim is:
Consider any Euclidean proof that there are infinitely many primes in some subset $S$ of
$\mathbb{Z}/N^*$. Then the set $S$ contains a complete division.
Since $\mathbb{Z}/N^*$ is abelian, the conjugacy part of the definition of abelianness doesn't come directly into this theorem, but it will arise in the proof.
So, what is a Euclidean proof? Well, at some step in the proof, I must build a prime $p$ and say "$p$ has the property $P$, and therefore $p$ is in $S$." In every Euclidean proof I have seen, $P$ is one of two things:
(1) There is a number $M$ such that $M$ is not in some subgroup $G$ of $(\mathbb{Z}/N)^*$. So $M$ has a prime divisor which is not in $G$, and we choose $p$ to be that prime factor. So an allowable step in a Euclidean-proof is showing that there are infinitely many primes not in some subgroup $G$ of $(\mathbb{Z}/N)^*$
(2) Some polynomial $f(X)$ factors in a particular way modulo $p$, and therefore $f$ lies in some class modulo $N$. (Generally, the observation is that $f$ has a root modulo $p$, but I'll allow more general things like $f$ has a quadratic factor" as well. For a polynomial $f$ of degree $n$, and a partition $\lambda$ of $n$, let $D(f, \lambda)$ be the set of primes such that the irreducible factors of $f$ have degrees $(\lambda_1 \lambda_2, \ldots, \lambda_r)$. And let $D(f, \lambda, N)$ be the image of $D(f, \lambda)$ modulo $N$. So an allowable step in an Euclidean-proof will be showing that there are infinitely many primes in some union of $D(f, \lambda, N)$'s.
So, what I will actually be proving is
Any subgroup $G$ of $(\mathbb{Z}/N)^*$, and any $D(f, n, \lambda)$ in $(\mathbb{Z}/N)^*$, is a
union of divisions.
So, suppose that $S$ is a set which does not contain any division, and let $T$ be disjoint from $S$, but contain an element representing every division class in $S$. If we have a Euclidean proof, it will construct some prime $p$. We may know that $p$ is not in various subgroups of $(\mathbb{Z}/N)^*$, or that $p$ is in some union of $D(f, \lambda, N)$'s. But that information can't distinguish whether $p$ is in $S$ or in $T$, so our proof can't show that $p$ is in $S$.
OK, my last boxed claim is a precise statement. Let's prove it.
It is easy to see that a subgroup of $(\mathbb{Z}/N)^*$ is a union of divisions (since we are in an abelian group, the conjugacy part of the definition is irrelevant). The complement of a subgroup is likewise such a union.
The interesting thing is $D(f, \lambda, N)$. Let $K$ be a Galois field where $f$ and $x^N-1$ both split. Let $G$ be the corresponding Galois group, so $G$ comes equipped with a map to $(\mathbb{Z}/N)^* \cong \mathrm{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})$.
Now, as you probably know, for $p$ a prime unramified in $K$, the factorization of $f$ modulo $p$ is determined by the Frobenius conjugacy class of $p$ in $G$. What you may or may not know is that it is actually determined by the division class of $p$! (For example, $x^5-1$ has the same factorization modulo primes which are $2 \mod 5$ and primes which are $3 \mod 5$.) To see this, just look at the recipe for reading the factorization off from the Frobenius class. It may or may not help to prove the following lemma: If $g \sim h$, and $X$ is a set with a $G$ action, then there is an order preserving bijection between the $g$ and the $h$ orbits in $X$.
So, the possible Frobenius classes in $G$ of primes in $D(f, \lambda)$ are unions of divisions (I am implicitly using the Cebotarov density theorem here). But then $D(f, \lambda, N)$ is just the projection to $(\mathbb{Z}/N)^*$ of the possible Frobenius classes in $G$ of primes in $D(f, \lambda)$. Since maps of groups take divisions to divisions, this shows that $D(f, \lambda, N)$ is a union of divisions.
Frobenius's density Theorem states that there are infinitely many primes with Frobenius in every division. (And, more precisely, that their Dirichlet density is the size of the division divided by the order of $G$.) It is significantly easier than Cebatarov's, using only the material from a first course in algebraic number theory and a first course in analytic number theory.
Cebatarov's density theorem is the "union" of Frobenius's and Dirichlet's theorems. What I am suggesting is that Euclidean methods, at best, can only get at their intersection.
I am not sure whether or not I think Euclidean proofs can get that far. If you think they can, give me a Euclidean proof that there are infinitely many primes which are in $\{ 3, 5 \}$ mod $7$. I can show infinitely many in $\{ 3,5,6 \}$, and infinitely many in $\{ 2,3,4,5 \}$, but I can't get the intersection.
Best Answer
Roughly speaking, the transference principle used in my work with Ben shows that the obstructions to solving (finite complexity) linear equations in dense sets of primes are the same as the obstructions to solving linear equations in dense sets of integers (modulo a technical issue involving dilating the equations by a medium-size integer $W$, coming from the "$W$-trick"). Thanks to the inverse conjecture for the Gowers norms, these obstructions are classified: local obstructions such as congruence classes are certainly obstructions, but in fact every nilsystem could potentially produce an obstruction.
For instance, the circle shift $x \mapsto x + \sqrt{2} \hbox{ mod } 1$ on ${\bf R}/{\bf Z}$ provides a family of obstructions based on the distribution of values of $\sqrt{2} n \hbox{ mod } 1$ in the set $S$ of interest. If for example $S$ happens to be contained in the Bohr set $\{ n: \{ \sqrt{2} n \} \in [0, 0.01] \}$ (where $\{x\} := x - \lfloor x \rfloor$ is the fractional part function), then there will be no patterns of the form $x, x+y, x+2y+2$ in the set $S$, despite the lack of congruence obstructions, simply because $\sqrt{2} x - 2 (\sqrt{2} (x+y)) + \sqrt{2}(x+2y+2) = 2\sqrt{2} \hbox{ mod } 1$ is too far away from zero. Intersecting this Bohr set with the primes would then give a positive density set of primes that had no patterns of the form $x,x+y,x+2y+2$.
For more complicated patterns one can cook up similar examples involving for instance the quadratic Bohr set $\{ n: \{ \sqrt{2} n^2 \}\in [0,0.01]\}$ or the more exotic ``bracket quadratic'' Bohr set $\{ n: \{ \lfloor \sqrt{2} n \rfloor \sqrt{3} n \} \in [0,0.01] \}$ that is associated to a Heisenberg nilflow. For a pattern of complexity $s$ (as defined in our paper, or by using the somewhat smaller notion of "true complexity" studied in later papers by Gowers-Wolf and others), every nilsystem of nilpotency class $s$ or less is a potential obstruction. (Congruence class obstructions can be thought of as corresponding to finite systems, which by convention can be viewed as "complexity 0" obstructions.)
On the other hand, if your set $S$ is equidistributed with respect to all such Bohr sets (after taking into account the local irregularities of distributions of the primes, for instance by using the "W-trick" of our paper), then the theory in my papers with Ben (together with one paper as well with Tamar Ziegler) will allow one to compute asymptotics for these patterns.