Let's denote:
$$F(a)=\frac{1}{2}+\frac{d}{da} \frac{S(a\pi)}{\pi^2} \tag{1}$$
Remember for later that $S(0)= \frac{\pi^2}{4}$.
From the second expression in the OP we obtain:
$$F(a)=a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n^2+a^2}} \tag{2}$$
In this answer I found that:
$$\frac{F(a)}{a}=\log 2 +\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k} \tag{3}$$
and
$$\frac{F(a)}{a}=\log 2 +\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx \tag{4}$$
Integrating (1) from $0$ to $a$, we have:
$$\int_0^a F(a') da'=\frac{a}{2}+\frac{S(a\pi)}{\pi^2}-\frac{1}{4}$$
On the other hand (3) gives us:
$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+2\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k+2}$$
This gives us explicit Taylor series for $S(a)$:
$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \frac{a^2}{2} \sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2 \pi} \right)^{2k} \tag{5}$$
While this is not a closed form, it still may prove useful.
From (4) we obtain:
$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^x-1)}dx$$
Which gives us an additional integral form for the function:
$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \pi a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^{\pi x}-1)}dx \tag{6}$$
