You are thinking too complicated. $\Re \bigl(\overline{z}f(z)\bigr) > 0$ on the unit circle implies that the function
$$g_\lambda(z) = (1-\lambda)z + \lambda f(z)$$
has no zeros on the unit circle for $\lambda \in [0,1]$:
$$\Re \bigl( \overline{z}g_\lambda(z)\bigr) = (1-\lambda)\lvert z\rvert^2 + \lambda \Re \bigl(\overline{z}f(z)\bigr) > 0$$
for $\lvert z\rvert = 1$. The number of zeros of $g_\lambda$ in the unit disk is therefore
$$N(\lambda) = \frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{g_\lambda'(z)}{g_\lambda(z)}\,dz = \frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \frac{(1-\lambda) + \lambda f'(z)}{(1-\lambda)z + \lambda f(z)}\,dz.$$
Since $g_0(z) = z$, we know $N(0) = 1$. We also know that $N(\lambda)$ depends continuously on $\lambda$.
Note that the argument is essentially Rouché's theorem. The condition $\lvert f-g\rvert < \lvert f\rvert$ in Rouché's theorem is just a simple way of stating a condition that ensures that $\lambda f(z) + (1-\lambda)g(z)$ has no zeros on the contour. $\Re \bigl(\overline{z}f(z)\bigr) > 0$ on $\lvert z\rvert = 1$ is another such condition.
Yes, the statement is true, and your solution is the standard argument. I'm not aware of a more elementary proof.
Let's write down the proof a little more systematically.
Since $f$ is by assumption injective (and analytic) on the boundary of the unit disk, $f\circ \gamma$ - where $\gamma$ is a positively oriented parametrisation of the unit circle - is a simple closed (continuously differentiable) curve, hence, by the Jordan curve theorem, it divides the plane into two disjoint domains whose common boundary is the trace of $f\circ\gamma$. Let $A_1$ denote the bounded domain (the interior of $f\circ\gamma$), and $A_2$ the unbounded domain (the exterior). Then the winding number
$$n(f\circ\gamma,z)$$
is constant on both domains, and $n(f\circ\gamma,z) = 0$ on $A_2$, while $n(f\circ\gamma,z) = \pm 1$ on $A_1$. But
$$n(f\circ\gamma,z) = \frac{1}{2\pi i}\int_{f\circ\gamma} \frac{dw}{w-z} = \frac{1}{2\pi i}\int_\gamma \frac{f'(\zeta)}{f(\zeta)-z}\,d\zeta$$
is, by the argument principle, the number of times $f$ attains the value $z$ in the open unit disk, hence non-negative, and it follows that $n(f\circ\gamma,z) \equiv 1$ on $A_1$. From $n(f\circ\gamma,z) \equiv 0$ on $A_2$, it follows that $f(D) \subset \overline{A}_1$, and by the open mapping theorem, it follows that $f(D)\subset A_1 = (\overline{A}_1)^{\Large\circ}$. Thus $f$ maps $D$ biholomorphically onto $A_1$.
Best Answer
As far as I am able to tell, your idea is correct. You can define $g$ by: $$ g(z) = \int_\gamma \frac{f'(z)}{f(z)} dz$$ because $\mathbb{D}$ is simply connected and the integral does not depend on the path. Or even more simply, just expand $\frac{f'(z)}{f(z)} = \sum_{n} a_nz^n$ and put $g(z) = \sum_{n} \frac{a_n}{n+1} z^{n+1}$. In any case, you find a holomorphic $g$ such that $g'(z) = \frac{f'(z)}{f(z)}$.
Put $h(z) = e^{g(z)}$; this is again holomorphic. Then $\frac{h'(z)}{h(z)} = \frac{f'(z)}{f(z)}$ by a simple computation. Let $r(z) := \frac{f(z)}{h(z)}$; one can compute that: $$ r'(z) = \frac{f'(z)h(z) - h'(z)f(z)}{h(z)^2} = 0 $$ Hence, $r(z)$ is a constant function. As a consequence, $f(z) = c h(z)$ for some constant $c$. Writing $c = e^{\alpha}$ (which is always possible) you get: $$ f(z) = e^{\alpha + g(z)} $$ so $\tilde{g}(z) := \alpha + g(z)$ is the sought function.