The first equation (or sequence of equations) follows from the second equation because $ x g''(x,t)+(1-x)g'(x,t)+t\frac{\partial g(x,t)}{\partial t} = 0$ where the prime indicates $\partial/\partial x$. You may just replace $g$ by $L_n$; in the last term, $t\cdot \partial/ \partial t$ picks the factor of $n$: just look how this operator acts on $t^n$.
The identity I wrote above is straightforward to prove. $-\frac{e^{-\frac{t x}{1-t}} t (1-x)}{(1-t)^2}+\frac{e^{-\frac{t x}{1-t}} t^2 x}{(1-t)^3}+t \left(\frac{e^{-\frac{t x}{1-t}}}{(1-t)^2}+\frac{e^{-\frac{t x}{1-t}} \left(-\frac{x}{1-t}-\frac{t x}{(1-t)^2}\right)}{1-t}\right) = 0$