At least in the first $r$-derivation, you missed some $\lambda$'s. Generally, always check your units if you derive solutation like that, as for exmple "$1+\frac{1}{r}$" just must be flawed.
And you'll solve your problem more directly using the Poisson equation $\Delta V(r)=\frac{\varrho(r)}{\varepsilon_0}$, and that with a $\Delta$ in explicitly spherical coordinates.
To take a short track here, your $V(r)$ looks like a Yukawa potential, i.e.
$$G(r)=\frac{\text{e}^{-\lambda r}}{4\pi\ r}$$
solves
$(\Delta-\lambda^2)G(r)=-\delta(\vec r)$ and hence
$$\Delta V(r)=4\pi A\left(-\delta(\vec r)+\frac{\lambda^2}{4\pi}\frac{\text{e}^{-\lambda r}}{r}\right)\equiv \frac{\varrho(r)}{\varepsilon_0}.$$
Integration over all of space, if you want to find the total charge. If you have a density like $\propto r^{-n}$, then clearly there are charges everywhere. Remark: For integration, notice that you're dealing with a three dimensional delta-function here (but from the Coulomb potential, you know that it gives you a single charge anyway). For the exp-integration, up to some small numbers, you can figure out the result just by power counting in $\lambda$'s.