I am not sure that is always true. Explore the case that $H(X)=H(Y)=1.$
When $n=2m$ is even , we should (I'd think) take $X=Y=\lbrace0,m\rbrace$ with equal weighting and then get $X+Y=Z$ with $H(Z)=1.$ I had mistakenly suggested that otherwise we should take $X=Y=\lbrace0,g\rbrace$ with equal probabilities to get $H(Z)=1.5$ However it was pointed to me that taking $X=Y=\lbrace-g,g,0\rbrace$ with probabilities $0.1135460976 , 0.1135460976 , 0.7729078048$ will yield $H(Z)=1.332599058$ in the case that $g=3$ and $n=9$ (or merely when $3g=n.$) By my calculations, that is optimal for a concentration on a subgroup of order $3$. Even when $Z$ has the 5 distinct members $\lbrace -2g,-g,0,g,2g \rbrace$, I get that $H(Z)=1.468267753$ which is a bit better than $1.5$.
However, in the case that $n=9$ it seems that taking $0$ with probability $0.8607538$ and each other 8 values with probability $0.017405778$ gives an even better entropy of about 1.27332346.
See the edit history for an earlier, faulty answer.)