EDIT: This argument claimed to prove something it did not, based on a stupid mistake (pointed out by Jacob Lurie in the comments).

Let's let ${\rm Pic}(KU)$ be the category of invertible $KU$-modules and weak equivalences, and $G$ be the subcategory of objects weakly equivalent to $KU$ or its suspension. These are symmetric monoidal with $\pi_0$ a group, so they are infinite loop spaces associated to spectra ${\rm pic}(KU)$ and $g$ respectively. $G$ acts on the category of $KU$-modules, and so your question is roughly: Can the map $\mathbb{Z}/2 = \pi_0 G \subset \pi_0 {\rm Pic}(KU)$ be lifted to a splitting of the map $BG \to B\mathbb{Z}/2$?

First we note that there is a fiber sequence $BGL_1(KU) \to {\rm Pic}(KU) \to \pi_0 {\rm Pic}(KU)$, and similarly for $G$. The map $BG \to B\pi_0 G$ is a principal bundle with fiber $BBGL_1(KU)$, obtained by delooping a fiber sequence of spectra $\Sigma {\rm gl}_1(KU) \to g \to H\pi_0(g)$; thus splitting is equivalent to asking for the triviality of the classifying map $B\pi_0(g) \to BBBGL_1(KU)$. If this were true, then the composite classifying map $B\pi_0(G) \to B^3 \pi_0(GL_1(KU))$ would also be trivial. This is a map $K(\mathbb{Z/2},1) \to K(\mathbb{Z/2},3)$, classifying the first stage in the Postnikov tower, and it is ~~either trivial or the nontrivial stable map classifying the Steenrod operation $Sq^2$~~ trivial.

We can use naturality to show that the stable map must be $Sq^2$. Let $\Sigma$ be the category of finite sets, which is the free symmetric monoidal category on one object; it has a symmetric monoidal map to the subcategory $G$ of $KU$-modules by sending the set with $n$ objects to $\Sigma^n KU$. Taking the associated spectra (K-theory objects) gives us a map from the sphere to $g$ which, on $\pi_0$ and $\pi_1$, are the projection map $\mathbb{Z} \to \mathbb{Z/2}$ and an isomorphism $\mathbb{Z/2} \to \mathbb{Z/2}$ (the twist map switches signs!) respectively. The k-invariants in the Postnikov tower are maps $H{\mathbb{Z}} \to \Sigma^2 H\mathbb{Z}/2$ and $H{\mathbb{Z}/2} \to \Sigma^2 H\mathbb{Z}/2$. The former is nontrivial (from Jacob's $\eta$-multiplication argument) and so naturality implies that the latter is nontrivial as well.

$KU$ could be any 2-periodic commutative ring spectrum with $2 \neq 0$ in $\pi_0(R)$ here.