No. A counterexample (essentially from Bourbaki's *Algèbre* VI.1 Exercice 12 a)) is the following.

We furnish $\mathbb{Z}$ with its usual structure of ordered group and consider the product of ordered groups $G=\mathbb{Z}^3$. This is a lattice, and for $(x,y,z),(u,v,w)\in G$ we have $$\textstyle\sup_G((x,y,z),(u,v,w))=(\sup(x,u),\sup(y,v),\sup(z,w)).$$ Now we consider the subgroup $H=\{(x,y,z)\in G\mid z=x+y\}$ of $G$, furnished with its induced structure of ordered group. This is also a lattice, as one readily checks that for $(x,y,x+y),(u,v,u+v)\in H$ we have $$\textstyle\sup_H((x,y,x+y),(u,v,u+v))=(\sup(x,u),\sup(y,v),\sup(x,u)+\sup(y,v)).$$ However, since $$\textstyle\sup_G((0,1,1),(1,0,1))=(1,1,1)\neq(1,1,2)=\sup_H((0,1,1),(1,0,1))$$ we see that $H$ is *not* a sublattice of $G$.