$a~\ddot{u}+b\left(\dot{u}\right)^2+\dot{u}+\dot{u}ce^u+e^u-e^{2u}+1=0$
$a\dfrac{d^2u}{dt^2}+b\left(\dfrac{du}{dt}\right)^2+(ce^u+1)\dfrac{du}{dt}+e^u-e^{2u}+1=0$
Case $1$: $a\neq0$
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0508.pdf
Let $v=\dfrac{du}{dt}$ ,
Then $\dfrac{d^2u}{dt^2}=\dfrac{dv}{dt}=\dfrac{dv}{du}\dfrac{du}{dt}=v\dfrac{dv}{du}$
$\therefore av\dfrac{dv}{du}+bv^2+(ce^u+1)v+e^u-e^{2u}+1=0$
$av\dfrac{dv}{du}=-bv^2-(ce^u+1)v+e^{2u}-e^u-1$
$v\dfrac{dv}{du}=-\dfrac{bv^2}{a}-\dfrac{(ce^u+1)v}{a}+\dfrac{e^{2u}-e^u-1}{a}$
This belongs to an Abel equation of the second kind.
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $v=\dfrac{1}{w}$,
Then $\dfrac{dv}{du}=-\dfrac{1}{w^2}\dfrac{dw}{du}$
$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=-\dfrac{b}{aw^2}-\dfrac{ce^u+1}{aw}+\dfrac{e^{2u}-e^u-1}{a}$
$\dfrac{dw}{du}=-\dfrac{(e^{2u}-e^u-1)w^3}{a}+\dfrac{(ce^u+1)w^2}{a}+\dfrac{bw}{a}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2