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.
A partial answer to (b): Consistently, yes. Löwenheim-Skolem implies that there are non-Archimedean real closed fields of cardinality $\aleph_1$, and it is consistent that
all sets of size $\aleph_1$ are of measure zero. (For example, this follows from MA plus non-CH.)
Alternatively: Take any model $V$ of set theory, let $K\in V$ be an uncountable
non-Archimedean real closed field, and add a Cohen real $c$ to $V$. In $V[c]$, the set of old reals has now measure zero, and $K$ is still a non-Archimedean real closed field.
I have a feeling that an absoluteness argument should now help to get the existence of $K$ in $V$, but I cannot get it to work. I do not even know if I can get a definable (say: analytic) $K$ in $V[c]$ (though it seems to me that I can get a $K$ containing a perfect set).
Best Answer
An interpretation of $(\mathbb R,+,\cdot)$ in $(\mathbb C,+,\cdot)$ in particular provides an interpretation of $\DeclareMathOperator\Th{Th}\Th(\mathbb R,+,\cdot)$ in $\Th(\mathbb C,+,\cdot)$. To see that the latter cannot exist in ZF:
The completeness of the theory $\def\rcf{\mathrm{RCF}}\rcf$ of real-closed fields is an arithmetical ($\Pi_2$) statement, and provable in ZFC, hence provable in ZF. Its axioms are clearly true in $(\mathbb R,+,\cdot)$, hence $\Th(\mathbb R,+,\cdot)=\rcf$.
Similarly, ZF proves completeness of the theory $\def\acfo{\mathrm{ACF_0}}\acfo$ of algebraically closed fields of characteristic $0$, hence $\Th(\mathbb C,+,\cdot)=\acfo$.
The non-interpretability of $\rcf$ in $\acfo$ is again an arithmetical statement ($\Pi_2$, using the completeness of $\acfo$), hence its provability in ZFC automatically implies its provability in ZF.
Of course, common proofs of some or all the results above may already work directly in ZF (e.g., if you take syntactic proofs of completeness, or if you make it work using countable models, etc.). My point is that it is not necessary to check the proofs, as the results transfer from ZFC to ZF automatically due to their low complexity.