= Solution
\b[Both assertions are false.] For <simple closure>, take the inclusion
$$
\mathcal M=(\{0\},+,-,0)\subseteq\mathcal N=(\mathbb Q,+,-,0).
$$
Both satisfy <DAG prime>, but the <quantifier-free formula> $x\ne0$ has a witness in $N$ and none in $M$. The same example works in the reduced language $\{+,0\}$.
For <quantifier elimination>, the <first-order sentence> $\exists x(x\ne0)$ distinguishes these two models. Every closed group term is zero, so every atomic closed equality is true in both models. Every <quantifier-free sentence>, being a Boolean combination of such equalities, has the same truth value in both. The distinguishing sentence therefore has no equivalent <quantifier-free sentence> modulo <DAG prime>. Hence
$$
\boxed{\mathrm{DAG}'\text{ does not have quantifier elimination}.}
$$
Including the trivial group is exactly what makes the proposed <simple closure> condition fail.
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