Consider the additive group $(\mathbb{R},+)$ and let $A$ be a subset
such that $\mathbb{Z}^++A\subseteq A$ (equivalently $1+A\subseteq A$,
and if and only if
$$
\cdots \subseteq 3+A \subseteq 2+A \subseteq 1+A \subseteq A
\subseteq -1+A \subseteq -2+A \subseteq -3+A \subseteq \cdots .
)$$
Then we have
$$
(*)\;\;\; (\mathbb{Z}^++A)\setminus (\bigcap_{i\in \mathbb{Z}} A+i)\subseteq
\mathbb{Z}^++(A\setminus (\mathbb{Z}^++A))
$$
Because, $1+A\subseteq A$ implies that $\mathbb{Z}^++A= 1+A$ and so
$$
(*)\Leftrightarrow$$ $$ (1+A)\setminus [(2+A)\cap(3+A)\cap(4+A)\cap \cdots]\subseteq [(1+A)\setminus
(2+A)]\cup [(2+A)\setminus (3+A)] \cup [(3+A)\setminus
(4+A)]\cup \cdots ,
$$
and the last one holds, clearly (even it is an equality).
Now, replacing $\mathbb{Z}^+$ by $\mathbb{Q}^+$ and $\mathbb{Z}$ by $\mathbb{Q}$, we would like to know that whether a similar relation to $(*)$ holds, that is: if $\mathbb{Q}^++A\subseteq A$, then is it true that $$ (*')\;\;\; (\mathbb{Q}^++A)\setminus (\bigcap_{i\in \mathbb{Q}} A+i)\subseteq \mathbb{Q}^++(A\setminus (\mathbb{Q}^++A))\;? $$
if no, can one find infinitely many such non-subsemigroup-subsets $A$ satisfying $(*')$ (and of course $\mathbb{Q}^++A\subseteq A$)?, and some examples for $$ (*'')\;\;\; \mathbb{Q}^++(A\setminus (\mathbb{Q}^++A)) \subsetneqq(\mathbb{Q}^++A)\setminus (\bigcap_{i\in \mathbb{Q}} A+i) \;? $$
Thanks in advance
Define $A=\mathbb Q^++\alpha\mathbb Z$, where $\alpha$ is any irrational. Since $\mathbb Q^++\mathbb Q^+=\mathbb Q^+$, this set satisfies $\mathbb Q^++A=A$. In particular, the right side of $(*')$ is just $\mathbb Q^+$. However, since every element of $A$ can be uniquely represented as $x+\alpha y$ where $x\in \mathbb Q^+$ and $y\in \mathbb Z$, $A+i$ consists only of those values for which $x>i$, and so $$\bigcap_{i\in \mathbb Q}A+i=\emptyset.$$ So, the left side of $(*')$ is $$(\mathbb Q^++A)\setminus \emptyset = \mathbb Q^++A=\mathbb Q^++\alpha\mathbb Z,$$ which is not a subset of $\mathbb Q^+$.
Remark. This sort of proof does not work for $\mathbb Z^+$; while the intersection over $i\in\mathbb Z$ is still empty, $A\setminus(\mathbb Z^++A)$ is rather large, which is enough to encompass the factor of the first set.