Suppose $S$ is a smooth surface birational to an abelian surface, and $f:S\to S'\subset\mathbb{P}^3$ is a birational morphism. Then $S'$ cannot have isolated singularities.
If it did, one could find a smooth hyperplane section $C\subset S^'$ missing the singular points. Since $V=\mathbb{P}^3- S'$ is smooth and affine, $H^i_c(V)=0$ for $i<3$ and so $H^1(S')=H^1(\mathbb{P}^3)=0$ by the exact sequence for $H_c$. On the other hand, let $U=S'-C\simeq S-D$ where $D=f^{-1}(C)$. Consider the long exact cohomology sequences for the pairs $(S,D)$ and $(S',C)$: $$ 0 \to H^1(S) \to H^1(U) \to H^2_D(S) \to\dots $$ and $$ 0 \to H^1(S') \to H^1(U) \to H^2_C(S') \to \dots $$ As $H^1(S')=0$, these imply that $H^1(S)$ injects into $H^2_C(S')$. But $C$ is irreducible, and contained in the smooth part of $S'$, so $H^2_C(S')$ is 1-dimensional. (Or you can argue with weights).