Proving without Zorn's Lemma: additive group of the reals is isomorphic to the additive group of the complex numbers

Unfortunately, there is no more elementary argument than going through some form of AC, because the result actually does depend on some amount of choice. As shown by e.g. C.J. Ash (see this 1973 J. Australian Math Society paper), an isomorphism between $(\mathbb{R},+)$ and $(\mathbb{C},+)$ implies the existence of a non-measurable set of reals. The paper has the full argument, but the short version is that (assuming that all sets of reals are measurable) one takes an isomorphism $f:\mathbb{R}\oplus\mathbb{R}\mapsto\mathbb{R}$, defines the sets $S_n=f[\mathbb{R}\oplus[n,n+1)]\cap(0,1)$ (that is, the image of $\mathbb{R}\oplus[n,n+1)$ under $f()$, intersected with the unit interval), and then shows that (a) the $S_n$ partition $(0,1)$ and (b) they all have the same measure. This is enough to contradict countable additivity.