Yes the reverse holds as well. This is really a definitional convention. We typically define "$f$ is integrable" to mean that both $\int_\Omega f^+ d\mu$ and $\int_\Omega f^- d\mu$ are finite (since, in the beginning, we define integration only for non-negative functions) then define $$\int_\Omega f d\mu = \int_\Omega f^+ d\mu - \int_\Omega f^- d\mu.$$ However, if both these are finite, one can easily see that $$\int_\Omega \lvert f \rvert d\mu = \int_\Omega f^+ d\mu + \int_\Omega f^- d\mu$$ and so $\int_\Omega \lvert f \rvert d\mu$ is finite as well.

Yes the reverse holds as well. This is really a definitional convention. We typically define "$f$ is integrable" to mean that both $\int_\Omega f^+ d\mu$ and $\int_\Omega f^- d\mu$ are finite (since, in the beginning, we define integration only for non-negative functions) then define $$\int_\Omega f d\mu = \int_\Omega f^+ d\mu - \int_\Omega f^- d\mu.$$ However, if both these are finite, one can easily see that $$\int_\Omega \lvert f \rvert d\mu = \int_\Omega f^+ d\mu + \int_\Omega f^- d\mu$$ and so $\int_\Omega \lvert f \rvert d\mu$ is finite as well.

Edit: @Vladamir has a much better answer in the comments. THe order in which we define things matters here. The value $\int_\Omega \lvert f \rvert d\mu$ is always defined (for measureable $f$). The value $\int_\Omega f \, d\mu$ is only defined when the other value is finite. Thus to check $f$ is integrable, we need to check $\int_\Omega \lvert f \rvert d\mu < \infty$ because a priori, the value $\int_\Omega f \, d\mu$ is not defined.