As already pointed out by @KStarGamer in the comment above, Mathematica can sum the recursion using RSolve[]. The result can be rewritten in terms of the regularized incomplete Euler beta function $I_x(a,b)$ and reads, for arbitrary $L_0=L(0)$ and $L_1=L(1)$, $$\tag{1}\label{eq:1} L(s)= 2^s s!\left[\frac{2 L_0-L_1}{\sqrt{2}} \left(1 - I_{-1}\left(s+1,\tfrac{1}{2}\right)\right) -(L_0 - L_1)\right]\,. $$

As already pointed out by @KStarGamer in the comment above, Mathematica can sum the recursion using RSolve[]. The result can be rewritten in terms of the regularized incomplete Euler beta function $I_x(a,b)$ and reads, for arbitrary $L_0=L(0)$ and $L_1=L(1)$, \begin{align} \tag{1a}\label{eq:1a} L(s) &= 2^s s!\left[\frac{2 L_0-L_1}{\sqrt{2}} \big(1 - I_{-1}\left(s+1,\tfrac{1}{2}\right)\big) -L_0 + L_1\right] \\ \tag{1b}\label{eq:1b} &= 2^s s!\left[\frac{2 L_0-L_1}{\sqrt{2}} I_{2}\left(\tfrac{1}{2},s+1\right) -L_0 + L_1\right] \end{align} According to Wikipedia,

The regularized incomplete beta function is the cumulative distribution function of the beta distribution, and is related to the cumulative distribution function $F(k;n,p)$ of a random variable $X$ following a binomial distribution with probability of single success $p$ and number of Bernoulli trials $n$: $$F(k;\,n,p)=\Pr \left(X\leq k\right)=I_{1-p}(n-k,k+1)=1-I_{p}(k+1,n-k).$$

This might link to the OP's original problem.

As already pointed out by @KStarGamer in the comment above, Mathematica can sum the recursion using RSolve[]. The result can be rewritten in terms of the incomplete Euler beta function and reads, for arbitrary $L_0=L(0)$ and $L_1=L(1)$, $$\tag{1}\label{eq:1} L(s)= 2^s \left[\frac{2 L_0-L_1}{\sqrt{2}} \left(s! - \frac{1}{2} \left(\tfrac{3}{2}\right)_s \,B_{-1}\left(s+1,\tfrac{1}{2}\right)\right) -(L_0 - L_1)\,s!\right]\,. $$ Note that the Pochhammer symbol $(a)_s=\Gamma(a+s)/\Gamma(a)$ was utilized to eliminate the $\sqrt\pi$.

As already pointed out by @KStarGamer in the comment above, Mathematica can sum the recursion using RSolve[]. The result can be rewritten in terms of the regularized incomplete Euler beta function $I_x(a,b)$ and reads, for arbitrary $L_0=L(0)$ and $L_1=L(1)$, $$\tag{1}\label{eq:1} L(s)= 2^s s!\left[\frac{2 L_0-L_1}{\sqrt{2}} \left(1 - I_{-1}\left(s+1,\tfrac{1}{2}\right)\right) -(L_0 - L_1)\right]\,. $$