I used the Cauchy Hadamard theorem to get the radius of convergence finding $r=\frac1e$. Now I have to check the convergence in $|x|=r$ but I didn't manage to prove the convergence or the divergence of the numerical series.
Moreover the exercise asks for the uniform convergence of the series that depends on the previous point.
For the case $x = \frac{1}{e}$ it is not difficult to show using the Taylor series expansion for $\log(1+x)$ that
$$\tag{*}\lim_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{e^n}= \frac{1}{e} \neq 0,$$
and, hence, the term limit with $x = -\frac{1}{e}$ does not exist since
$$\limsup_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{(-e)^n}= \frac{1}{e},\quad \liminf_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{(-e)^n}= -\frac{1}{e}$$
and the series does not converge at the endpoints $x = \pm\frac{1}{e}$
The series in question is a power series and, hence, it converges on any compact subset of the open interval $\left(-\frac{1}{e}, \frac{1}{e}\right)$.
That leaves the question of uniform convergence of the series on the entire open interval $\left(-\frac{1}{e}, \frac{1}{e}\right)$.
To show that the series fails to converge uniformly on an open interval with $\frac1e$ as an endpoint, it is enough to show that the terms fail to converge uniformly to $0$. This is indeed the case because with
$$f_n(x) = \left(1+\log\left(1+\frac1n\right)\right)^{n^2}x^n$$
and $x_n = \frac{\left(1-\frac1n\right)}{e} \in (0,\frac1e)$ we have
$$\lim_{n \to \infty} f_n(x_n) = \lim_{n\to \infty}\frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{e^n}\left( 1 - \frac1n\right)^n= \frac{1}{e^2}\neq 0$$
To prove (*), consider
$$\log a_n =\log \frac{\left(1+\log\left(1+\frac1n\right)\right)^{n^2}}{e^n} = n^2 \log\left(1+\log\left(1+\frac1n\right)\right)-n\\ = n^2 \log \left(1+\frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3} \right)\right)- n\\= n^2 \left[\left(\frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3} \right)\right)- \frac{1}{2}\left(\frac{1}{n} - \frac{1}{2n^2} + \mathcal{O}\left(\frac{1}{n^3} \right)\right)^2 + \mathcal{O} \left(\frac{1}{n^3} \right) \right]-n\\ = -1 + \mathcal{O}\left( \frac1n\right) $$
The RHS converges to $-1$ as $n \to \infty$ and, thus, $a_n \to e^{-1}$ as $n \to \infty$.