When $a>2$, this sum cannot be written as an Euler product $\prod_{p\text{ prime}} (\sum_{k\ge 0} a_{p^k} p^{-ks})$. We'll prove this by using the following fact: a Dirichlet series has an Euler product if and only if the corresponding arithmetic function is multiplicative ($f(nm)=f(n)f(m)$ if $(n,m)=1$) - prove this by expanding the Euler product!
First case: $a=b=1$. It is well-known that $\mu(n)$ is multiplicative (as an inverse of a multiplicative function), hence
$$(*) \sum_{n \ge 1} \frac{\mu(n)}{n^s} = \prod_{p} (\sum_{k \ge 0} \mu(p^k) p^{-ks})=\prod_{p} (1-p^{-s})$$
Alternatively, since $\mu$ is the inverse of the constant function $\zeta = 1$ which has the representation $\sum_{n \ge 1} n^{-s} = \prod_{p} (1-p^{-s})^{-1}$, one can deduce $(*)$ by $(\sum_{n \ge 1} n^{-s}) (\sum_{n \ge 1} \mu(n) n^{-s})=1$.
Second case: $a=2,b=1$. The function $f(n)=\mu(n) \cdot 1_{2 \nmid n} = \begin{cases} 0 && 2|n \\ \mu(n) && 2 \nmid n \end{cases}$ is multiplicative, because both $\mu(n)$ and $1_{2 \nmid n} = \begin{cases} 0 && 2 |n \\ 1 && 2\nmid n \end{cases}$ are.
Thus, we can expand the Dirichlet series of $f$ as follows:
$$\sum_{n \ge 0} \frac{\mu(an+b)}{(an+b)^s} = \sum_{n \ge 0} \frac{\mu(2n+1)}{(2n+1)^s} = \sum_{m \ge 1} \frac{f(m)}{m^s}=$$
Here comes the Euler product:
$$\prod_{p} (\sum_{k \ge 0} f(p^k) p^{-ks})= (\sum_{k \ge 0} f(2^k) 2^{-ks}) \prod_{p \neq 2} (\sum_{k \ge 0} \mu(p^k) p^{-ks}) = \prod_{p \neq 2} (1-p^{-s})$$
Third case: $a>2$. Sadly, $f(n)=\mu(n) \cdot 1_{n \equiv b \mod a}$ is no longer multiplicative. To show this, I'll assume Dirichlet's theorem on primes in arithmetic progressions. It states that if $n,m$ are coprime positive integers, then there are infinitely many primes satisfying $p \equiv m \mod n$.
Thus we can construct 2 distinct primes $p_1, q_1$ such that $p_1 \equiv 1 \mod a, p_2 \equiv b \mod a$. We have $f(p_1 q_1) = \mu(p_1 q_1) 1_{p_1 q_1 \equiv b \mod a} = 1$, and $f(p_1) f(q_1) = \mu(p_1)\mu(q_1) 1_{p_1 \equiv b \mod a}1_{q_1 \equiv b \mod a}= 1_{p_1 \equiv b \mod a, q_1 \equiv b \mod a}$.
Thus, $f(p_1 q_1)=f(p_1)f(q_1)$ implies $b \equiv 1 \mod a$. Now we'll construct 2 additional distinct primes $p_2, q_2$ such that $p_2, q_2 \equiv a-1 \mod a$. Since $p_2 q_2 \equiv 1 \mod a$, yet $p_2 \neq 1 \mod a$, we have $f(p_2)f(q_2) = 0$ yet $f(p_2 q_2)=1$.
It follows that $f$ isn't multiplicative.
Now we'll express our Dirichlet series as a linear combinations of Dirichlet series' with Euler product. In other words, we want to write $\mu(n) \cdot 1_{n \equiv b \mod a}$ as a linear combination of multiplicative functions.
We'll use characters. Characters modulo $m$ are functions from the positive integers to the complex numbers, satisfying 3 properties:
They are periodic with period $m$: $\chi(a)=\chi(a+m)$.
They are (completely) multiplicative: $\chi(ab) = \chi(a) \chi(b)$.
If $(a,m) >1$, $\chi(a)=0$. Otherwise, $\chi(a) \neq 0$.
The theory of characters modulo $m$ is very well understood. For any $m$ you can construct distinct $\phi(m)$ characters. Given the factorization of $m$, it is a direct procedure to construct those characters. Read more here. There's a lot of interesting theory there, and explicit examples of characters.
Exercise 1: Show that $\chi(1)=1$ for any character, and that $\chi(a)$ is always a root of unity of order dividing $\phi(m)$.
Exercise 2: If $\chi$ is a character, $\overline{\chi}$ (complex conjugation) is also a character. $\overline{\chi}(a) = \chi(a^{-1})$, where $a^{-1}$ denotes a positive integer satisfying $a^{-1} \cdot a \equiv 1 \mod m$. Show that $\overline{\chi}(a) = \begin{cases} 0 && (a,m) > 1 \\ \frac{1}{\chi(a)} && (a,m)=1 \end{cases}$.
Exercise 3: The product of 2 characters is also a character.
Enough with the exercises. Suppose $\chi_{i}, 1 \le i \le \phi(m)$ are my distinct characters.
Theorem: Suppose $a$ is coprime to $m$.
$$\frac{1}{\phi(m)} \sum_{i=1}^{\phi(m)} \overline{\chi_i(a)}\chi_i(n) = 1_{n \equiv a \mod m}$$
The proof is actually not hard. If $(n,m)>1$ both sides are 0. Otherwise, by exercise 2, this identity is equivalent to
$$\sum_{i=1}^{\phi(m)} \chi_i(x) = \phi(m) 1_{x \equiv 1 \mod m}$$
where $x \equiv \frac{n}{a} \mod m$.
One direction is easy - if $x=1 \mod m$, all the characters evaluated on $x$ give the value 1.
The other direction is very nice - assume $x \neq 1 \mod m$. Let $S(x)=\sum_{i=1}^{\phi(m)} \chi_i(x)$. Use exercise 3 to show that $\{ \chi_i \}_{i} = \{ \chi_i \chi_j \}_{i}$, and deduce that $S(x) \chi_j(x) = S(x)$, for any $j$. Since the order of $x$ modulo $m$ is not $1$, one can find a character that doesn't give the value $1$ on $x$ (I can't elaborate without using more group theory), so for some $j$ we can cancel $\chi_j(x) - 1$ in the equality $S(x)(\chi_j(x) - 1)=0$ and get that $S(x)=0$. $\blacksquare$
With this theorem at hand, the decomposition is straightforward: we can write $\mu(n) \cdot 1_{n \equiv b \mod a}$ as $\frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} \chi_i(n)\mu(n)$, and this shows:
$$\sum_{n \ge 0} \frac{\mu(an+b)}{(an+b)^s} = \sum_{n \ge 1} \frac{\mu(n) 1_{n \equiv b \mod a}}{n^s}=$$
$$\frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} (\sum_{n \ge 1} \frac{\mu(n) \chi_i(n)}{n^s})$$
And each summand has an Euler product, since $\mu(n) \chi_i(n)$ is multiplicative (a product of 2 multiplicative functions):
$$\sum_{n \ge 1} \frac{\mu(n) \chi_i(n)}{n^s} = \prod_{p \text{ is prime}} (\sum_{k \ge 0} \mu(p^k)\chi_i(p^k)p^{-ks})=\prod_{p \text{ is prime}} (1-\chi_i(p)p^{-s})$$
You might recognize this as the inverse of the so-called L-function $L(s, \chi_i)$, defined by:
$$L(s, \chi_i) = \sum_{n \ge 1} \frac{\chi_i(n)}{n^s} = \prod_{p \text{ is prime}} (1-\chi_i(p)p^{-s})^{-1}$$
Thus, we can write your function as $$\frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} L(s,\chi_i)^{-1}$$
When $a=b=1$ or $a=2,b=1$ (the 2 cases discussed at the beginning), there is exactly one character, since $\phi(2)=\phi(1)=1$, and this sum becomes one element: $L(s, \chi)^{-1}$ where $\chi$ is the "trivial" character - the trivial character modulo $m$ is 1 when the value is coprime to $m$ and $0$ otherwise.
Hope this was readable. I can't help but notice that you ask many questions of this kind, reading chapter 16 of this classc will help you.
EDIT: In general,
$$\sum_{k \ge 0} \frac{f(ak+b)}{(ak+b)^s} = \frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} h(s,\chi_i)$$
where $h(s,\chi_i) = \sum_{n \ge 1} \frac{f(n)\chi_i(n)}{n^s}$, which has an Euler product expansion when $f$ is multiplicative. This show that you can express this kind of sums with at most $\phi(a)$ functions with Euler product expansions.
I can give a partial answer to this. My apologies if I am telling you what you already know.
First, let's review how the explicit formula for the Merten's function is derived. We have that
$$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}.$$
We have
$$M(x) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \frac{1}{\zeta(s)} \frac{x^s}{s} dx,$$
and the residue theorem gives you the explicit formula you mentioned above. Note that the zeros of $\zeta$ become poles of $\zeta^{-1}$, so the two summations in your formula correspond to the nontrivial zeros and the trivial zeros of Riemann zeta function respectively. Also the $1/s$ term gives us a pole at $s=0$, giving us a residue of $1/\zeta(0) = -2$.
More generally, we can define a function $d_k(n)$ by the Dirichlet series given by $\zeta^k$:
$$\zeta(s)^k = \sum_{n=1}^\infty d_k(n) n^{-s}.$$
Then, for $k\geq 1$, $d_k(n)$ gives the number of representations of $n$ as a product of $k$ positive integers, given explicitly by your product formula above. In particular, $d_2(n)$ gives the number of positive divisors of $n$. Also, as you mentioned, $d_{-1}(n) = \mu(n)$, the Möbius function.
(For $k<-1$, I am have not checked whether my definition using the Dirichlet series will coincide with your product formula.)
Now you can attempt to find an explicit formula by carrying out the contour integral
$$D_k(x) = \sum_{n<x} d_k(n) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \zeta(s)^k \frac{x^s}{s} dx.$$
Now we can see that the explicit formula you will get will differ greatly depending on whether $k$ is positive or negative. If $k$ is positive, the main term will come from the residue of the pole at $s=1$, and this main term will be of the form $xP(\log x)$, where $P$ is a polynomial of degree $k-1$. If $k$ is negative, you'll pick up residues at all the zeros of $\zeta$, as in the case of $k=-1$.
Best Answer
This may help: Gevorg Hmayakyan wrote a short paper which gives a recursive formula for the $\mu(n)$.
https://ghmath.files.wordpress.com/2010/06/mobius.pdf
It's not for the faint of heart, however. But maybe a faster algorithm could come out of it.