Hint: the torsion group of $G$ is the group whose elements are those elements of $G$ which have finite order. They form a subgroup of $G$. The identity element is by definition, in this torsion group.
The torsion group may very well contain only the identity: the trivial subgroup, in that event.
ADDED: to confirm your comment/question/edit:
Yes, the torsion subgroup of $\mathbb Z \times (\mathbb Z/n\mathbb Z)$ is $0 \times (\mathbb Z/n\mathbb Z)$.
And as Pete L. Clark suggests in his comment below: "To get a proof, just take it systematically: can you first show that every element that you've written down has finite order? That's almost obvious. Now write down any element other than yours and show that it does not have finite order."