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 a solution of the first order autonomous ODE
$$u' = \frac{1- \Psi_+(u )+\Psi_-(u )}{2}$$ with $u(0)=0$. Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u$ solve the initial problem.
Rmk 1. 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 $F$.
Rmk 2. Since $\Psi_\pm$ are (local) inverses of $\Phi$ at its minimum point $b$, they are not Lipschitz at $0$; therefore the equation for $u$ may fail to have uniqueness!