Your other metric is not a solution to [Einstein field equation][1]. The metric coefficients $g_{00}$ and $g_{rr}$ are related to each other through the equation below ($u\equiv r^2/R^2$). Using it one can check whether the new metric satisfies it. 
$${\rm e}^{-\lambda}\frac{\rm d}{{\rm d} u} \left( {\rm e}^{-\lambda}\frac{\rm d}{{\rm d}u}{\rm e}^{\nu}\right) = \frac{1}{4}\frac{\rm d}{{\rm d}u}\left(\frac{1-{\rm e}^{-2\lambda}}{u}\right)~{\rm e}^{\nu}~.\tag{1}$$

The easiest way to do it is to notice that both metrics satisfy ${\rm e}^{\nu}={\rm e}^{-\lambda}$ what simplifies the equation $(1)$ to  
\begin{equation}\label{compactform2}
\frac{\rm d}{{\rm d} u} \left(\frac{\rm d}{{\rm d}u}{\rm e}^{\nu}\right) = \frac{1}{2}\frac{\rm d}{{\rm d}u}\left(\frac{1-{\rm e}^{-2\lambda}}{u}\right),~\tag{2}
\end{equation}
The solution of equation $(2)$ reads $${\rm e}^{-2\lambda}=1+\frac{C_{1}}{\sqrt{u}}+C_{2}~u\equiv1+\frac{\tilde{C}_{1}}{r}+\tilde{C}_{2}~r^2.~\tag{3}$$
The vacuum solution (Schwarzschild solution) corresponds to $\tilde{C_{2}}\equiv0$. 

Concluding, the only metric satisfying $~g_{00}\cdot g_{rr}=1$ relation is the Schwarzschild metric (with or without cosmological term).

[1]:https://physics.stackexchange.com/a/679431/281096