Here is a supporting evidence for the conjecture made in the end. Let $f$ be a plynomial
(or an entire function). We have $f'=1-P$; where $P$ is a power series with positive coefficients. I claim that the zero of $f'$ which is closest to the origin
is positive. Indeed this zero is the closest singularity to the origin of the function
$$1/f'=1/(1-P)=1+P+P^2+P^3+... .$$
But this is a power series with positive coefficients, therefore the
closest singularity to the origin must be on the positive ray by Pringsheim's theorem.

However, this does not prove the conjecture, even for polynomials, because there can be other
obstacles to injectivity, different from a zero of $f'$. One has to prove that these other
obstacles cannot occur for such series.

EDIT: In fact the conjecture is true! Let $f(z)=z-Q(z)$, $Q$ has positive coefficients and
double zero at zero. Suppose without loss of generality that the zero of $f'$ of
the smallest modulus is outside the disc $D$ of radius $r$ with center at the origin.

We wish to prove that $f$ is injective in this disc. Consider $P=Q'$. Then $P(0)=0$ and $P$ has positive coefficients. So $P$ is increasing on $(0,r)$ and cannot take the value $1$. Thus
$|P(z)|$ is less than $k$ in $D$, with some $k\in(0,1)$, because coefficients are positive.

Now we prove the statement by contradiction. Suppose $f(z_1)=f(z_2)$ with $z_j$ in the unit disc. Integrating along the straight line
segment connecting $z_1$ and $z_2$, we obtain
$$0=f(z_2)-f(z_1)=\int_{z_1}^{z_2}f'(\zeta)d\zeta=z_2-z_1-\int_{z_1}^{z_2}P(\zeta)d\zeta.$$
Estimating the integral, we obtain
$$|z_1-z_2|=\left|\int_{z_1}^{z_2}P(\zeta)d\zeta\right|\leq k|z_1-z_2|,$$
which is a contradiction. This proves your conjecture.