I propose you take $$g(x):=\frac{N}{c}\exp\frac{-x}{N}.$$ I too doubt that you can find an answer independent on $N$.

The method I used to determine $g$ might lead to an answer which is more intreseting, so I provide it here. Define $$\phi(a):=\int_0^\infty\frac{\exp(-x)}{(1+ag(x))^N} dx$$ so that $$\phi'(a)=\int_0^\infty u'(x)v(x) dx $$ with $$u(x)=\frac{1}{a(1+ag(x))^N} \\ v(x)=\frac{g(x)\exp(-x)}{g'(x)}.$$Integrating by parts gives $$a\phi'(a)=\alpha\phi(a)-\frac{g(0)}{g'(0)(1+ag(0))^N} $$ if we have the relation $$v'(x)=-\alpha\exp(-x) , ~~~~\alpha\in\mathbb C$$ which holds whenever $$ g(g'+g'')=(\alpha+1)(g')^2 .$$ Assume for now that this equation is satisfied. Then $$ \phi(a)=-a^\alpha \int_0^a \frac{g(0)}{g'(0)(1+tg(0))^Nt^{1+\alpha}} dt $$ which is of the required growth-type only if $\alpha=-N$, in which case $$\phi(a)\sim_\infty -\frac{\ln a}{a^Ng'(0)},$$ so that $$c=-\frac{1}{g'(0)}.$$

Now we solve for $g$, which satisfies the equation $$1+\frac{g''}{g'}=(1+\alpha)\frac{g'}{g}$$. This equation is integrable and there exists two constants $\delta, \gamma$ such that $$g(x)=(\gamma+\delta\exp(-x))^{-1/\alpha}.$$ The constant $\gamma$ is useless, while $\alpha$ and $\delta$ are completely determined.