Solution using minimal calculus: Let $y=1-x$ and $z=\lfloor{\log_2(\frac{1}{1-x})}\rfloor$.
Lemma $1$:
$x^{2^{n}} \geq (1-2^{n-1}y),\; \forall n \in \mathbb{N^+}$.
Proof:
We proceed by induction. For the base case $n=1$, we have $1 \geq 1-y$. Suppose this is true for $n$. Then, assuming $1-2^{n}y \geq 0$, $x^{2^{n+1}}=\left(x^{2^{n}}\right)^{2}$ $\geq \left(1-2^{n}y\right)^{2} = 1-2^{n+1}y+2^{2n}y^{2} \geq 1-2^{n+1}y$. If $1-2^{n}y < 0$, then this is obviously true.
We have $$\log_2(1-x)+\sum_{i=1}^\infty x^{2^{i}} \\= \sum_{i=1}^\infty x^{2^{i}}-\log_2\left(\dfrac{1}{1-x}\right) \\\geq \sum_{i=0}^z (1-2^{i}y)-\log_2(1-x)\\= \left\lfloor{\log_2\left(\frac{1}{1-x}\right)}\right\rfloor+1-\log_2\left(\frac{1}{1-x}\right)-\left(2^{z+1}-1\right)y \geq -2.$$
It is also well known that $x^{2^{z+1}} \leq \frac{1}{e}$, so $$\sum_{i=1}^\infty x^{2^{i}}+\log_2(1-x) \\= \sum_{i=1}^\infty x^{2^{i}}-\log_2\left(\dfrac{1}{1-x}\right) \\\leq \sum_{i=1}^z x^{2^{i}}+\sum_{i=1}^\infty x^{i2^{z+1}}-\log_2\left(\dfrac{1}{1-x}\right) \\\leq \left\lfloor {\log_2\left(\dfrac{1}{1-x}\right)}\right\rfloor-\log_2(\dfrac{1}{1-x})+\dfrac{1}{e-1} \\\leq \dfrac{1}{e-1}$$