Let $\mathbb{Z}^*$ be the set of all elements of $\mathbb{Z}$ cancellable under integer multiplication. Let $(z_1,z_2){\boxtimes}(z_3,z_4){\iff}z_1*z_4=z_2*z_3$. Let $[[z_1,z_2^*]]_{\boxtimes}+[[z_3,z_4^*]]_{\boxtimes}=[[z_1z_4^*+z_3z_2^*,z_2^*z_4^*]]_{\boxtimes}$.
How to prove that for all $[[z_1,z_2^*]]_{\boxtimes}$, $[[z_1,z_2^*]]_{\boxtimes}+[[0,a^*]]_{\boxtimes}=[[z_1,z_2^*]]_{\boxtimes}$, where $a^*$ is any element of $\mathbb{Z}^*$?
By direct computation we have:
$[[z_1,z_2^{\ast}]]_{\boxtimes} + [[0,a^{\ast}]]_{\boxtimes} = [[z_1a^{\ast} + 0z_2^{\ast},z_2^{\ast}a^{\ast}]]_{\boxtimes}$
To show that $[[0,a^{\ast}]]_{\boxtimes}$ is indeed an (additive right-) identity for $(\Bbb Z \times \Bbb Z^{\ast})/\sim_{\boxtimes}$ we must then show the following are $\boxtimes$-equivalent: that is,
$[[z_1a^{\ast} + 0z_2^{\ast},z_2^{\ast}a^{\ast}]]_{\boxtimes} \sim_{\boxtimes} [[z_1,z_2^{\ast}]]_{\boxtimes}$, which means showing:
$(z_1a^{\ast} + 0z_2^{\ast})\ast(z_2^{\ast}) = (z_2^{\ast}a^{\ast})\ast z_1$.
Computing the LHS, we have:
$(z_1a^{\ast} + 0z_2^{\ast})\ast(z_2^{\ast}) = (z_1a^{\ast})\ast (z_2^{\ast})$
$= z_1z_2^{\ast}a^{\ast}$, which is clearly equal to the RHS. Note the multiplication here refer to "ordinary" integer multiplication, which is commutative, and associative.
By commutivity of $+$ in $(\Bbb Z \times \Bbb Z^{\ast})/\sim_{\boxtimes}$ (verfiy this!), the class $[[0,a^{\ast}]]_{\boxtimes}$ is indeed thus an additive identity. Traditionally, we take $a^{\ast} = 1$, since the choice is arbitrary (this works out well when we want to embed a copy of $\Bbb Z$ in $(\Bbb Z \times \Bbb Z^{\ast})/\sim_{\boxtimes}$).