For simplicity, let's assume initially that the support of $\mu$ is the whole real line, so that $F$ is a homeomorphism $ \mathbb{R} \to (0,1) $. Let's denote $b:=F(0)=\mu(-\infty,0]=1-\mu[0,+\infty)\in(0,1)$.
The relevant function is the strictly convex function $\Phi :(0,1)\to\mathbb{R}$ defined as $$\Phi(t):=\int_{b}^t F^{-1}(s)\, ds $$ Finally, let's denote $\Psi_{+}:=(\Phi_{|[b,1)})^{-1}$ and $\Psi_-:=(\Phi_{|(0,b]})^{-1}$, and consider the solution $u:[0,+\infty)\to \mathbb{R}$ of the first order autonomous ODE
$$u'(t) = \frac{1- \Psi_+(u )+\Psi_-(u )}{2},\quad t\ge0$$ with $u(0)=0$, given explicitly by $$u^{-1}(x)=2\int_0^x\frac{dy}{1-\Psi_+(y)+\Psi_-(y)}$$ (note that the integrand is well defined, positive and increasing).
Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u$ solve the initial problem written as $$ 2\phi_+ F'(\phi_+)\phi_+'=2\phi_- F'(\phi_-)\phi_-'=1-F(\phi_+)+F(\phi_-)$$ that we can write equivalently, because $\phi_\pm=F^{-1}\circ F\circ \phi_\pm=\Phi'\circ F\circ \phi_\pm$, in the form $$ 2(\Phi\circ F\circ \phi_+)'=2(\Phi\circ F\circ \phi_-)'=1-\Psi_+\circ\Phi\circ F\circ \phi_+ +\Psi_-\circ\Phi\circ F\circ \phi_- .$$
Conversely, if $\phi_\pm$ solve the initial system, the latter equation shows that $\Phi\circ F\circ \phi_+=\Phi\circ F\circ \phi_-$ because they coincide at $0$ and have the same derivative, and in fact solve the above equation for $u$.
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Rmk. The assumption on the support of $\mu$ is not crucial; if it is not the whole real line, $F$ is constant on some open set, and $F^{-1}$ has jumps and it is only defined as a left inverse of $F$; $\Phi$ is however well defined. It may also be defined as a Legendre transform of an antiderivative of $F$.