Since $\varphi$ is continuous and doesn't vanish we can take $F: J \longrightarrow\mathbb{R}$ to be the antiderivative of $1/\varphi$. $(F \circ f)' (x) = 1, \forall x \in I \Rightarrow F(f(x))=x+a$ $(F \circ g)' (x) = 1, \forall x \in I \Rightarrow F(f(g))=x+b$ Now, because $0 \in Im(f-g)$ we get that $a=b$. $F'$ doesn't vanish, which means $\text{sgn}(F'(x))$ is constant, so $F$ is strictly monotonic, which implies $F$ is injective. Since $F(f(x))=F(g(x))$ and $F$ is injective, we conclude that $f=g$