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.)

I now think that it is quite possibly true. I explored 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 $1.5$.

In the case that $n=9$ it seems that taking $0$ with probability $0.8607538$ and each other 8 values with probability $0.017405778$ is best if we want only two values, but this gives (once calculations are done correctly....) an entropy of about 1.5917.

I still wonder what is best for $n=121$ (or some other composite $n$ with no small factors)

See the edit history for an earlier, faulty answer.)

I'd guess that perhaps they can be supported on $G$ but that it is not always the case that they must be. Suppose that the entropy for both $X$ and $Y$ should be $1$. Then indeed if $n=2m$ we should (I'd think) take $X=Y=\lbrace0,m\rbrace$ with equal weighting and then get $X+Y=Z.$ But for $n$ odd (and composite) I'd think that $X$ and $Y$ should each be $\lbrace0,g\rbrace$ for some $g \ne 0$ with $X+Z$ supported on the set $\lbrace 0,g,2g \rbrace.$ with probabilities $1/4,1/4$ and $1/2.$ Of course we can also shift and have $X=\lbrace0,g\rbrace$ and $Y=\lbrace0,-g\rbrace.$ In any case, there is no gain (nor loss) in confining $g$ to a proper subgroup.

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.)