This is to give an explicit description of the domain of convergence of your second series provided in the answer by Alexandre Eremenko.
In the $(s,t)$-plane, the domain is the interior of the set
\begin{equation}
\{(s,t)\in\mathbb R^2\colon s<g(t)\},
\end{equation}
where
\begin{equation}
g(t):=\inf_{a\in(0,1)}h(a,t)
\end{equation}
and
\begin{equation}
h(a,t):=\frac{(1-a) \ln (1-a)+(1+a) \ln (1+a)-(1-a) t}a.
\end{equation}
The partial derivative of $h(a,t)$ in $a$ is
\begin{equation}
\frac{t-\ln(1-a^2)}{a^2}.
\end{equation}
So, if $t>0$, then $h(a,t)$ is increasing in $a\in(0,1)$ from $h(0+,t)=-\infty$ and hence $g(t)=-\infty$.
If $t<0$, then $h(a,t)$ attains its minimum in $a\in(0,1)$ when $a$ is the root of the equation $t-\ln(1-a^2)=0$, that is, when $a=\sqrt{1-e^t}$. Thus, the domain of convergence for your second series is
\begin{equation}
\{(s,t)\in\mathbb R^2\colon t<0\ \&\ s<h(\sqrt{1-e^t},t)\}.
\end{equation}
Part of this region (for $t>-8$) is shown here:
Note that $h(\sqrt{1-e^t},t)\to\ln4=1.386\dots$ as $t\to-\infty$.