Please see below for my partial proof. I'm stuck at the point where I've shown $a+b=a'+nl+b'+nk$. If someone could help show how to complete this proof that would be great.
Suppose $[a],[b]\in\mathbb{Z}$ and $[a]=[a']$ and $[b]=[b']$. Aice adds $[a]$ and $[b]$ as $[a]+[b]=[a+b]$. Bob adds them as $[a']+[b']=[a'+b']$. Show that their answers are the same.
$Proof.$ Since $[a]=[a']$ we know that $a\equiv a'\pmod n$ and thus $a-a'=nl$ for some $l\in\mathbb{Z}$. Similarly since $[b]=[b']$ we know that $b\equiv b'\pmod n$ and thus $b-b'=nk$ for some $k\in\mathbb{Z}$. So then $a=a'+nl$ and $b=b'+nk$.
Then
$a+b=a'+nl+b'+nk$
Since $a+b = a'+b' + (l+k)n$, we have $a+b = (a'+b') \pmod n$ and so $a+n \sim a'+b'$.