This is not a complete solution but it shows, for a simplified version of the problem, how a frequenty used approach works.
Consider the simplified problem to calculate the sum
$$s(k,x) = \sum _{n=0}^{\infty } \frac{P_n(x)}{k+n}$$
Where $k>0$ is a real parameter.
The generating function for the Legendre polynomials is
$$g(t,x) = \frac{1}{\sqrt{t^2-2 t x+1}} = \sum _{n=0}^{\infty } t^n P_n(x)$$
Now writing
$$\frac{1}{k+n}=\int_0^{\infty } \exp (-z (k+n)) \, dz$$
inserting this into the expression for $s$ we get
$$\sum _{n=0}^{\infty } P_n(x) \int_0^{\infty } \exp (-z (k+n)) \, dz$$
Interchanging summation and integration gives
$$\int_0^{\infty } \left(\sum _{n=0}^{\infty } P_n(x) \exp (-z (k+n))\right) \, dz$$
Extracting the factor $ \exp (- k z)$ from the sum this can be written as
$$\int_0^{\infty } \exp (- k z) \sum _{n=0}^{\infty } \exp (- n z) P_n(x) \, dz$$
Now the sum can be done using the formula for the generating function of the Legendre polynomials (with $t = \exp(- z))$ with the result
$$\sum _{n=0}^{\infty } \exp (-n z) P_n(x) = \frac{1}{\sqrt{-2 x e^{-z}+e^{-2 z}+1}}$$
Hence we arrive at this integral representation of $s$:
$$s1(k,x) = \int_0^{\infty } \frac{e^{-k z}}{\sqrt{-2 e^{-z} x+e^{-2 z}+1}} \, dz$$
Mathematica gives for this integral the closed form expression:
$$s1(k,x) = \frac{1}{k (k+1) (k+2)}\left( (k+1) (k+2) F_1\left(k;-\frac{1}{2},-\frac{1}{2};k+1;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)+2 k (k+2) x F_1\left(k+1;\frac{1}{2},\frac{1}{2};k+2;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)-k (k+1) F_1\left(k+2;\frac{1}{2},\frac{1}{2};k+3;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)\right)$$
Here $F_1$ is the Appell1 hypergeometric function (http://mathworld.wolfram.com/AppellHypergeometricFunction.html).
Admittedly, this is not a very "pleasant" expression, but the result for the complete problem of the OP can be expected to be even uglier, if it has a closed form at all, which I doubt.
EDIT
For integer $k$ we get closed form expressions (always for the simplified Problem). The first few are
$$s(0,x) = \log (2)-\log \left(-x+\sqrt{2-2 x}+1\right)$$
Notice that for $k=0$ the n-sum starts at $n=1$.
$$s(1,x) = \log \left(\frac{x-\sqrt{2-2 x}-1}{x-1}\right)$$
$$s(2,x) = \sqrt{2-2 x}+2 x \coth ^{-1}\left(\sqrt{2-2 x}+1\right)-1$$
$$s(3,x) = \frac{1}{2} \left(\left(6 x^2-2\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)$$
$$s(4,x) = \frac{1}{6} \left(6 x \left(5 x^2-3\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+5 x \left(3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)-2 \sqrt{2-2 x}+4\right)$$
We consider the sequence $\{P_n(t)\}$ of Legendre polynomials. We describe how to construct a generating function
\begin{align*}
G(t,u)=\sum_{n=0}^\infty P_n(t)u^n
\end{align*}
and how to derive the region of convergence. We closely follow example 7.4 from Asymptotics and Special Functions by F.J.W. Olver.
We recall Rodrigues' formula
\begin{align*}
P_n(t)=\frac{(-1)^n}{2^nn!}\frac{d^n}{dt^n}\left\{\left(1-t^2\right)^n\right\}
\end{align*}
and get using Cauchy's integral formula for the $n$-th derivative of an analytic function Schläfli's integral
\begin{align*}
P_n(t)=\frac{1}{2^{n+1}\pi i}\int_{\mathcal{C}}\frac{(z^2-1)^n}{(z-t)^{n+1}}dz
\end{align*}
in which $\mathcal{C}$ is any simple closed contour that encircles $z=t$; here $t$ may be real or complex. For fixed $\mathcal{C}$ and sufficiently small $|u|$, the series
\begin{align*}
\sum_{n=0}^\infty\frac{(z^2-1)^nu^n}{2^{n+1}\pi i(z-t)^{n+1}}
\end{align*}
converges uniformly with respect to $z\in\mathcal{C}$, by the M-test. By integration and summation we obtain
\begin{align*}
\frac{1}{2\pi i}\int_{\mathcal{C}}\left\{1-\frac{(z^2-1)u}{2(z-t)}\right\}^{-1}\frac{dz}{z-t}=\sum_{n=0}^\infty P_n(t)u^n=G(t,u).
\end{align*}
It follows
\begin{align*}
G(t,u)=-\frac{1}{\pi i}\int_{\mathcal{C}}\frac{dz}{uz^2-2z+(2t-u)}=-\frac{1}{u\pi i}\int_{\mathcal{C}}\frac{dz}{(z-z_1)(z-z_2)}
\end{align*}
where
\begin{align*}
z_1=\frac{1-\sqrt{1-2tu+u^2}}{u},\qquad z_2=\frac{1+\sqrt{1-2tu+u^2}}{u},
\end{align*}
We observe if $u\to 0$, then $z_1\to t$ and $|z_2|\to\infty$. Hence for sufficiently small $|u|$, $\mathcal{C}$ contains $z_1$ but not $z_2$. The residue theorem yields
\begin{align*}
G(t,u)=-\frac{2}{u}\frac{1}{z_1-z_2}=\frac{1}{\sqrt{1-2tu+u^2}}
\end{align*}
We conclude, the desired expansion is given by
\begin{align*}
\color{blue}{\frac{1}{\sqrt{1-2tu+u^2}}=\sum_{n=0}^\infty P_n(t)u^n}\tag{1}
\end{align*}
provided that $|u|$ is sufficiently small and the chosen branch of the square root tends to $1$ as $u\to 0$.
For $t\in[-1,1]$ the singularities of the left-hand side of (1) both lie on the circle $|u|=1$, hence in this case the radius of convergence of the series on the right-hand side is unity.
Best Answer
Note that this is a solution as an integral, not a closed form solution
Let's start with the generating function
$$ \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^\infty P_n(x)t^n $$
Multiply by $t^{k-1}$ and get
$$ \frac{t^{k-1}}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^\infty P_n(x)t^{n+k-1} $$
We replace $t\mapsto u$ and integrate $u$ from $0$ to $t$ and get
$$ \int_0^t\frac{u^{k-1}}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{n+k} $$
Now divide by $t$, do another substitution $t\mapsto v$ and integrate this from $0$ to $t$ to get
$$ \int_0^t\frac{1}{v}\int_0^v\frac{u^{k-1}}{\sqrt{1-2xt+t^2}}\,dudv=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^2} $$
This double integral is performed over a triangular region in $uv$ space. Switching the order of integration yields
$$ \int_0^t\int_u^t\frac{1}{v}\frac{u^{k-1}}{\sqrt{1-2xt+t^2}}\,dvdu=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^2} $$
We can evaluate the integral in $v$ to get
$$ \int_0^t\frac{u^{k-1}\ln(t/u)}{\sqrt{1-2xt+t^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^2} $$
We continue this process, but the result is weird. We again divide by $t$, replace $t\mapsto v$, and integrate from $0$ to $t$ to get
$$ \int_0^t\frac{1}{v}\int_0^v\frac{u^{k-1}\ln(v/u)}{\sqrt{1-2xu+u^2}}\,dudv=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3} $$
We swap the order and factor some terms to make the equation a bit simpler to get
$$\int_0^t\left(\int_u^t\frac{\ln(v/u)}{v}\,dv\right)\frac{u^{k-1}}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3} $$
We do a change of variable; let $w=v/u$ and $dw=(1/u)dv$ to get
$$\int_0^t\left(\int_1^{t/u}\frac{\ln(w)}{w}\,dw\right)\frac{u^{k-1}}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3} $$
Integrating with respect to $w$ gives
$$\frac{1}{2}\int_0^t\frac{u^{k-1}\ln^2(t/u)}{\sqrt{1-2xu+u^2}}\,du=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^3} $$
This process can be repeated indefinitely (I'll skip the rigorous details, since they are long and tedious) and we are left with the generalized equation of
$$ \frac{1}{(\alpha-1)!}\int_0^t\frac{u^{k-1}\ln^{\alpha-1}(t/u)}{\sqrt{1-2xu+u^2}}=\sum_{n=0}^\infty\frac{P_n(x)t^{n+k}}{(n+k)^\alpha},\;\;\;\;\alpha=1,2,3,... $$
Let $t=1$ and we have
$$\boxed{ \frac{(-1)^{\alpha-1}}{(\alpha-1)!}\int_0^1\frac{u^{k-1}\ln^{\alpha-1}(u)}{\sqrt{1-2xu+u^2}}=\sum_{n=0}^\infty\frac{P_n(x)}{(n+k)^\alpha},\;\;\;\;\alpha=1,2,3,... }$$
We can see how a solution for $\alpha=1$ could be obtained, since the log term disappears in this form. A more straightforward approach to this answer comes from a well known result of
$$\int_0^1 t^\mu\ln^k(t)\,dt=(-1)^k\frac{k!}{(\mu+1)^{k+1}}$$
Thus, we could have simply multiplied the generating function by $t^{k-1}\ln^{\alpha-1}(t)$, integrated from $0$ to $1$, then rearranged a bit.