I proved that for every $0<a<1$, the series $ \sum_{n=0}^\infty \left( \frac{x^{2n+1}}{2n+1} - \frac{x^{n+1}}{2n+2} \right) $ converges uniformly at $[-a,a]$ to $\frac{1}{2}\log (x+1)$.
Now, if we apply $x=1$ we get $\log 2$, but $\log 2 \ne \frac{1}{2} \log 2$.
Why isn't it a contradiction to Abel's theorem?
There is no contradiction, since in Abel's theorem, you consider a power series of the form: $$ \sum_{n=0}^{\infty}a_nx^n $$ Here, you don't have a power series, due to the powers $x^{2n+1} \neq x^{n+1}$ in the summand of your initial series.
Hoping to be clear.