How does the union of two machines which accept language of form $0^{mx+b}$ look like
It is written union of machines (I suppose here machine means finite automaton), but it is only an union of languages, i.e. the union of the languages accepted by said automata.
Indeed it should be finite otherwise it could be not regular. To prove this fact, for $k\in \mathbb{N}$, consider $L_k = \{0^{n(2k+3)},n\in\mathbb{N}\}$, and $L = \bigcup\limits_{k\in\mathbb{N}}L_k$. Then $L$ is the set of words over the alphabet $\{0\}$ that have a length which is the multiple of an odd number greater than $2$. That means that the complement of $L$ is $\overline{L} = \{0^{2^n}, n\in\mathbb{N}\}$. Since $\overline{L}$ is not even context-free, then $L$ cannot be regular.
$L = \{0^{2n}, n\in \mathbb{N}\}\cup \{0^{3n}, n\in\mathbb{N}\}$ satisfy the condition. To prove it, suppose $L = \{0^{an+b}, n\in\mathbb{N}\}$. Since $\varepsilon \in L$, then it means that $b = 0$. Since $0^2\in L$, it means that $a\mid 2$. Since $0^3\in L$, it means that $a\mid 3$. We conclude that $a\mid\text{gcd}(2, 3) = 1$, so that $a = 1$. Since it is clear that $L \neq \{0^n, n\in\mathbb{N}\}$ (because $0^5\notin L$ for example), $L$ is regular and not of the said form.
There seems to be a bit of misunderstanding about the variable $x$ here… And to be precise and avoid any mistake, you should write $\{0^{2x+3}, x\in\mathbb{N}\}$ instead of just $0^{2x+3}$. That way you will understand that $x$ is just a free variable and it does not matter how you write it.