Let $K$ be a knot in a closed, oriented 3-manifold $Y$. It is a standard fact that if $K$ is (at least rationally) null-homologous, then $H_1(Y-K;\mathbb{Z})$ is isomorphic to $H_1(Y;\mathbb{Z})\oplus \mathbb{Z}$. I'm curious about the case where $K$ is not rationally nullhomologous, i.e. when $[K]$ has infinite order in $H_1(Y)$.
What is $H_1(Y-K)$ if $K$ is not rationally null-homologous?
To make a few initial observations, let $N(K)$ denote a tubular neighborhood of our knot $K$ (which is assumed to satisfy $[K]\neq 0$ in $\mathbb{Q}$-homology) and let $\mu$ be a meridian of $K$ in $\partial N(K)$. Then:
by the "half lives, half dies" principle, $\mu$ must represent a trivial element in $H_1(Y-K;\mathbb{Q})$ and thus a finite-order element in $\mathbb{Z}$-homology;
$Y$ is obtained from $Y- \mathring{N}(K)$ by attaching a 2-handle along $\mu$ (and capping off the remaining $S^2$-boundary with a 3-ball), so its first homology should be the quotient of $H_1(Y-K)$ by the subgroup generated by $\mu$.
Based on this, my guess is that $H_1(Y-K)\cong H_1(Y) \oplus \mathbb{Z}/n$ where $n$ is the order of $\mu$.
All your steps are correct, you actually get $H_1(Y-K)/(\mu) \cong H_1Y$. Now in fact $[\mu]=0$ in your case, cf. Homology of knot exterior in general manifold for not null-homologus knot
In case you are missing any details let me know. (the only thing you basically use is Poincaré duality and some differential topology).