It seems not.

It was shown by Di Prisco and Todorcevic (and reproved later by at least three sets of authors) that if sufficiently large cardinals exist (e.g., a proper class of Woodin cardinals), then after forcing with $\mathcal{P}(\omega)/\mathrm{Fin}$ (the infinite subsets of $\omega$, ordered by mod-finite containment) to produce a selective ultrafilter $U$, there is no selector (i.e., set meeting each equivalence class in exactly one point) for the equivalence relation $E_{0}$ (mod-finite equivalence on $\mathcal{P}(\omega)$) in the inner model $L(\mathbb{R})[U]$.

It also seems to follow from ZF + DC$_{\mathbb{R}}$ (which holds in $L(\mathbb{R})[U]$) that the existence of a discontinuous homomorphism from either of $(\mathbb{R}, +)$ or $(\mathbb{C}, +)$ to itself implies the existence of an $E_{0}$ selector, as we will show below. Since a discontinuous automorphism of $(\mathbb{C}, +, \times)$ restricts to one for $(\mathbb{C}, +)$, this answers the question.
The proof is the same for each of $(\mathbb{R}, +)$ and $(\mathbb{C}, +)$; moreover, the existence of each type of homomorphism implies the existence of the other. I haven't tried writing it up this way, but it seems that the argument can be carried out over an arbitrary complete additive metric group satisfying the triangle inequality. The existence of a discontinuous homomorphism of $(\mathbb{R}, +)$ easily gives one for
$(\mathbb{C}, +)$; we give a proof of the reverse direction at the end of this answer.

So, let $h$ be a discontinuous homomorphism from $(\mathbb{R}, +)$ (or $(\mathbb{C}, +)$) to itself.
As shown in the proof of Theorem 1 of a 1947 paper by Kestelman, for each positive real number $\delta$, $h$ is unbounded on $\{ x : |x| < \delta \}$. The same proof shows that the same fact holds for $(\mathbb{C}, +)$ (moreover, the fact follows easily from the definition of "discontinuous homomorphism"). Applying DC$_{\mathbb{R}}$, we may find $\{ x_{i} : i < \omega \}$ such that (1) each $|x_{i}|$ is more than $\sum \{ |x_{j}| : j > i\}$
and such that (2) for each $i$, $|h(x_{i})| - \sum \{ |h(x_{j})| : j < i \} > i.$

Let $X = \{ x_{i}: i < \omega \}$ and let $Y$ be the set of reals (or complex numbers) which are sums of (finite or infinite) subsets of $X$ (note that all the infinite sums converge).
By condition (1) on $X$, each $y \in Y$ is equal to $\sum \{ x_{i} : i \in S_{y}\}$ for a unique subset $S_{y}$ of $\omega$.
Let $F$ be the equivalence relation on $Y$ where $y_{0} F y_{1}$ if and only if $S_{y_{0}}$ and $S_{y_{1}}$ have finite symmetric difference.
By condition (2) on $X$, the $h$-preimage of each bounded subset of $\mathbb{R}$ ($\mathbb{C}$) intersects each $F$-equivalence class in only finitely many points (since if the bounded set is contained in an interval of length $i$, then for every $y$ in the intersection $S_{y} \setminus i$ is the same, which can be seen be consideration of the maximum point of disagreement between the sets $S_{y}$). It follows then that there is an $F$-selector : for each equivalence class, let $n \in \mathbb{Z}^{+}$ be minimal such that the $h$-preimage of $[-n, n]$ intersects the class, and then pick the least element of this intersection. Since $Y/F$ is isomorphic to $\mathcal{P}(\omega)/E_{0}$ via the map $y \mapsto S_{y}$, there is then an $E_{0}$-selector.

As for getting a discontinuous homomorphism of $(\mathbb{R}, +)$ from one on $(\mathbb{C}, +)$ : Suppose that $h$ is a homomorphism of $(\mathbb{C}, +)$. Define $f_{0},\ldots,f_{3}$ on $\mathbb{R}$ as follows:
(1) If $h(x) = a + bi$, then $f_{0}(x) = a$.
(2) If $h(x) = a + bi$, then $f_{1}(x) = b$.
(3) If $h(iy) = a + bi$, then $f_{2}(y) = a$.
(4) If $h(iy) = a + bi$, then $f_{3}(y) = b$.
Then each of $f_{0},\ldots,f_{3}$ is a homomorphism of $(\mathbb{R}, +)$.
Since $h(x + iy) = h(x) + h(iy) = f_{0}(x) + if_{1}(x) + f_{2}(y) + if_{3}(y),$ if all of $f_{0},\ldots,f_{3}$ are continuous
then $h$ is.