Yes. Your involution is a field automorphism, and since it is of order $2$, its fixed field $F$ is a subfield of $\mathbb C$ such that $[\mathbb C:F]=2$. (This incidentally implies that $F$ is a real-closed field, hence the condition on sums is automatically satisfied, you don’t have to assume it.) In particular, both $F$ and $\mathbb C\setminus F$ have cardinality $2^{\aleph_0}$, hence you can define $\phi$ as follows. Let $X$ be a selector on the equivalence relation given by $a\sim a'$ on $\mathbb C\setminus F$. Take any bijection of $F$ and $\mathbb R$, any bijection of $X$ and $\{z\in\mathbb C:\operatorname{Im}(z)>0\}$, and put $\phi(a)=\phi(a')^*$ for $a\in\mathbb C\setminus(F\cup X)$. You can make it slightly more constructive by observing that $\mathbb C=F(i)$, and $(a+ib)'=a-ib$ for $a,b\in F$. Thus, if $\phi$ is a bijection of $F$ and $\mathbb R$, we can extend it to an involution-preserving bijection $\mathbb C\to\mathbb C$ by putting $\phi(a+ib)=\phi(a)+i\phi(b)$.

We can also make $\phi$ preserve addition ($F$ is a vector space over $\mathbb Q$ of dimension $2^{\aleph_0}$, hence it is isomorphic to $\mathbb R$ as a vector space). However, we cannot in general make it preserve multiplication as well: $F$ is not necessarily isomorphic to $\mathbb R$ as a field, for example it may be non-archimedean. In fact, if $F$ is *any* real-closed field of cardinality $2^{\aleph_0}$, then $F(\sqrt{-1})$ is an algebraically closed field of characteristic 0 and transcendence degree $2^{\aleph_0}$, and as such it is isomorphic to $\mathbb C$. We may thus obtain an involutive field automorphism of $\mathbb C$ whose fixed field is isomorphic to $F$.