You can subtract $\frac{1}{x}$ and multiply by $xe^{-x/2}$ to avoid the Cauchy product and instead use the known series expansions of $e^x$. You would get $$e^{x/2} -e^{-x/2} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}$$
Write out the series expansion of the left hand side to get $$\sum_{n=0}^{\infty} \frac{x^n}{2^n n!} - \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n n!} \stackrel?= \frac{1}{2} \sum_{n=1}^\infty\frac{(-1)^{n-1}x^{n+1}}{n!2^{n-1}}\sum_{k=0}^{\lfloor(n-1)/2\rfloor}\frac1{2k+1} + \sum_{n=0}^\infty\frac{(-1)^nx^{n+1}}{n!2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac1{2k+1}$$
From here, you can just show that the $x^n$ terms are the same on both sides. It's clearly equal for $n = 0$ and $n = 1$ ($[x^0] = 0$ and $[x^1] = 1$). Other than that, you would need to show the relation $$\frac{1}{2^n n!} - \frac{(-1)^n}{2^n n!} \stackrel?= \frac{1}{2}\frac{\left(-1\right)^{n-2}}{\left(n-1\right)!2^{n-2}}\sum_{k=0}^{\lfloor (n-2)/2 \rfloor}\frac{1}{2k+1}+\frac{\left(-1\right)^{n-1}}{\left(n-1\right)!2^{n-1}}\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}\frac{1}{2k+1}$$
Multiplying by $2^n n!$ and factoring the RHS, this simplifies to $$1 - (-1)^n \stackrel?= 2n\left(-1\right)^{n}\left( \sum_{k=0}^{\lfloor (n-2)/2 \rfloor}\frac{1}{2k+1}-\sum_{k=0}^{\lfloor (n-1)/2 \rfloor}\frac{1}{2k+1}\right)$$
Splitting this up into even and odd cases makes it easier. If $n$ is even, then the LHS is $0$ and $\lfloor (n-2)/2 \rfloor = \lfloor (n-1)/2 \rfloor$, so the RHS is $0$ as well. If $n$ is odd, then the LHS is $2$. The RHS would be $$-2n\left( -\frac{1}{2\lfloor (n-1)/2 \rfloor + 1} \right) = -2n \cdot \frac{-1}{n} = 2$$
Therefore, the left hand side and the right hand side are both equal.
Here a possible alternative to the approach outlined in the comments. Let us assume we start with the FL-expansion of $\text{Li}_2(x),\log^2(1-x)$ or $\log(x)\log(1-x)$ (they all are reasonably simple and related to each other via the dilogarithm reflection formula), for instance
$$\text{Li}_2(x) = (\zeta(2)-1)+\sum_{n\geq 1}\frac{2n+1}{n^2(n+1)^2}\,P_n(2x-1).$$
We have
$$ \frac{\text{Li}_2(x)}{x}= \zeta(3)+\sum_{n\geq 1} c_n\,P_n(2x-1) $$
and by Bonnet's recursion formula
$$ x P_n(2x-1) = \frac{n+1}{4n+2}P_{n+1}(2x-1)+\frac{1}{2}P_n(2x-1)+ \frac{n}{4n+2}P_{n-1}(2x-1) $$
so the coefficients of the FL-expansion of $\frac{\text{Li}_2(x)}{x}$ are given by the solution of the recurrence
$$ \frac{n}{4n-2} c_{n-1} + \frac{1}{2} c_n + \frac{n+1}{4n+6}c_{n+1} = \frac{2n+1}{n^2(n+1)^2}\tag{R} $$
with the initial conditions $c_0=\zeta(3)$ and $c_1=\pi^2-6-3\zeta(3)$. My version of Mathematica is not able to directly crack this through $\text{RSolve}$, but we may establish once again the superiority of humans over machines through insights, i.e. our chaotic thought process.
By eye-balling it is not difficult to get
$$ c_n=(2n+1)\int_{0}^{1}\frac{\text{Li}_2(x)}{x}P_n(2x-1)\,dx = (-1)^n(2n+1)\left[\zeta(3)-H_n\frac{\pi^2}{3}+r(n)\right]$$
with $r(n)\in\mathbb{Q}$, and since $c_n\to 0$ we may expect that the structure of $r(n)$ is close to the one of $2 H_n H_n^{(2)}-H_n^{(3)}$. If we let $c_n=(-1)^n (2n+1) d_n$ in $(R)$ we end up with
$$ -n d_{n-1} + (2n+1) d_n - (n+1) d_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R'} $$
and by letting $e_n=d_n-d_{n-1}$, then $f_n = ne_n$, we get
$$ n e_n - (n+1) e_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R''} $$
$$ f_n - f_{n+1} = 2(-1)^n \left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\tag{R'''} $$
so
$$ f_n = \sum_{k\geq n}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$
$$ e_n = \frac{1}{n}\sum_{k\geq n}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$
$$ d_n = \sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right) $$
and by dropping the $\zeta(3)$-related part
$$\boxed{ c_n = (-1)^n(2n+1)\sum_{m=1}^{n}\frac{1}{m}\sum_{k\geq m}2(-1)^k \left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right). }$$
A similar approach can be used for finding the FL-expansions of
$\text{Li}_n(x)$, since
$\text{Li}_{n+1}(x)=\int\text{Li}_n(x)\frac{dx}{x}$ and $\int
P_n(x)\,dx=\frac{1}{2n+1}(P_{n+1}-P_n)$. According to my knowledge the
FL-expansion of $\text{Li}_4$ is yet unpublished, but I think it is
not a big deal to directly share this technique with the community (I hope Marco agrees :/)
Best Answer
One view for the series: $$ S_{2}(x)=\sum_{n=0}^\infty \frac{\psi ^{(0)}\left(\frac{n+3}{2}\right)-\psi ^{(0)}\left(\frac{n+2}{2}\right)}{2(n+1)}\,x^{2n+1}. $$ Using an integral definition for $\psi(x)$ then $$ \psi ^{(0)}\left(\frac{n+3}{2}\right) - \psi^{(0)}\left(\frac{n+2}{2}\right) = 2 \, \int_{0}^{1} \frac{u^{n+1} \, du}{1+u}. $$ With this the series reduces to $$ S_{2}(x) = - \frac{1}{x} \, \int_{0}^{1} \frac{\ln(1-x^2 \, u) \, du}{1+u}. $$ This can be evaluated further and is $$ S_{2}(x) = \frac{1}{x} \, \left( \text{Li}_{2}\left(\frac{1}{1+x^2}\right) - \text{Li}_{2}\left(\frac{1-x^2}{1+x^2}\right) - \ln(1-x^2) \, \ln\left(\frac{2 \, x^2}{1+x^2}\right) \right). $$
An example of the use of this result is: $$ S_{2}(1) = \text{Li}_{2}\left(\frac{1}{2}\right) = \frac{\zeta(2) - \ln^{2}(2)}{2}. $$
Using integration by parts on the integral for $S_{2}(x)$ leads to \begin{align} - x \, S_{2}(x) &= \int_{0}^{1} \frac{\ln(1-x^2 \, u) \, du}{1+u} \\ &= \left[ \ln(1+u) \, \ln(1-x^2 \, u) + x^2 \, \int \frac{\ln(1+u) \, du}{1 - x^2 \, u} \right]_{0}^{1} \\ &= \left[ \ln(1+u) \, \ln(1-x^2 \, u) - \ln(1+u) \, \ln\left(\frac{1-x^2 \, u}{1+x^2}\right) - \text{Li}_{2}\left(\frac{(1+u) \, x^2}{1+x^2}\right) \right]_{0}^{1} \\ &= - \left( \text{Li}_{2}\left(\frac{2 \, x^2}{1+x^2}\right) - \text{Li}_{2}\left(\frac{x^2}{1+x^2}\right) - \ln 2 \, \ln(1+x^2) \right) \end{align} which gives $$ S_{2}(x) = \frac{1}{x} \, \left( \text{Li}_{2}\left(\frac{2 \, x^2}{1+x^2}\right) - \text{Li}_{2}\left(\frac{x^2}{1+x^2}\right) - \ln 2 \, \ln(1+x^2) \right). $$ With this form there may be a transformation that can be used that could reduce $S_{2}(x)$ further. It does not appear that it could be reduced to $\text{Li}_{2}(x)$.