(where $a>0$ and $b>0$)
I proved that if $\frac ab$ is rational then $a\Bbb Z+b\Bbb Z = \frac an\Bbb Z= \frac bm\Bbb Z$ where $\frac nm=\frac ab$ and $\gcd(n,m)=1$ and is therefore not dense. However I'm blocked on the case where $\frac ab$ is irrational
Let $r=\inf\{\,x\in a\Bbb Z+b\Bbb Z\mid x>0\,\}$. Clearly, $0\le r\le a<\infty$. Then either $r\in a\Bbb Z+b\Bbb Z$ or there exists a strictly decreasing sequence in $a\Bbb Z+b\Bbb Z$ converging to $r$.
Assume there are $n_1,n_2,m_1,m_2$ with $r<an_1+bm_1<an_2+bm_2<2r$. Then $0<a(n_2-n_1)+b(m_2-m_1)<r$, contradicting the definition of $r$ as infimum. Hence for $r>0$, the case of a strictly decreasing sequence in $a\Bbb Z+b\Bbb Z$ converging to $r$ is not possible. We conclude that either $r=0$ or $0<r\in a\Bbb Z+b\Bbb Z$.
Assume $0<r\in a\Bbb Z+b\Bbb Z$, say $r=an+bm$. If $an'+bm'\notin r\Bbb Z$, then there exists $k\in\Bbb Z$ with $kr<an'+bm'<(k+1)r$, so $0<a(n'-kn)+b(m'-km)<r$, again contradicting the definition of $r$ as infimum.
We conclude $r=0$. Then for every $\epsilon>0$, there exists $x\in a\Bbb Z+b\Bbb Z$ with $0<x<\epsilon$. Now if $(u,v)\subset \Bbb R$ is any open interval, pick $x\in a\Bbb Z+b\Bbb Z$ with $0<x<v-u$. Then $x\Bbb Z$ intersects $(u,v)$ as was to be shown.
You already proved that if $\frac ab$ is rational, then $a\mathbb Z + b\mathbb Z$ is not dense.
Conversely, assume that $U := a\mathbb Z+b\mathbb Z$ is not dense in $\mathbb R$. Let $$ \rho:= \inf\{u\in U\,|\, u>0\}. $$ Since $U$ is not dense, we have $\rho>0$.
But then I claim that $U = \rho \mathbb Z$. Indeed, $\rho\mathbb Z\subseteq U$ is clear. Given $u\in U$, there is some $k\in \mathbb Z$ with $k\rho \le u < (k+1)\rho$. Then $0\le u-k\rho < \rho$ and $u-k\rho \in U$ implies $u-k\rho=0$ by the definition of $\rho$. Hence, $u = k\rho\in \rho\mathbb Z$.
It follows that $\mathbb Z = \frac a{\rho}\mathbb Z + \frac b{\rho}\mathbb Z$. In particular, $\frac{a}{\rho}, \frac{b}{\rho}\in \mathbb Z$., and hence $\frac ab = \frac{a/\rho}{b/\rho}$ is rational.
