Surfaces in $\mathbb{P}^3$ with isolated singularities It is classically known that every smooth, complex, projective surface $S$ is birational to a surface $S' \subset \mathbb{P}^3$ having only ordinary singularities, i.e. a curve $C$ of double points, containing a finite number of pinch points and a finite number of triple points, which are triple also for $S'$. The proof is obtained by embedding $S$ in $\mathbb{P}^5$ and by taking a projection $$\pi_{L} \colon S \to \mathbb{P}^3,$$
where $L \subset \mathbb{P}^5$ is a general line. This is the method originally used by M. Noether in order to prove his famous formula $$\chi(\mathcal{O}_S)=\frac{1}{12}(K_S^2+c_2(S)),$$
see Griffiths-Harris Principles of Algebraic Geometry, p. 600.

Question. Is it also true that every smooth, complex, projective surface $S$ is birational to a surface $S' \subset \mathbb{P}^3$ having only isolated singularities? And if not, what is a counterexample?

 A: 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).
Remark: as indicated below this argument is false (and any cohomological argument along the same lines runs into the same problem).
A: IMPORTANT EDIT 12-2015
There is this paper of Tokunaga "Irreducible Plane Curves of Albanese Dimension Two" which based on cited work of Kulikov constructs surfaces in P^2 with isolated singularities and Albanese dimension two.  In other words, the conjecture that for a smooth model of a normal in P^2 the Albanese dimension should be one is false.  This makes the answer to the question of the OP open. This information was given to me by R. Gurjar.
Original Answer- 
To the best of my knowledge this is a long standing open problem.  I cannot recall a reference, as this is something I studied in the 1980's, but I recall this being phrased as an unsolved problem from the 19th century Italian school.  The conjecture is that no normal surface in P^3 is birational to a smooth surface which has two dimensional image in it's Albanese.  One specific case of this that has been studied more extensively are Zariski surfaces:z^n = f(x,y) where f is a polynomial of degree n with only cusps and nodes as singularities. There are lots of information about when such a surface is irregular, but beyond that not much is known.  I believe that even if f is a sextic polynomial it is unknow whether or not the resulting surface can have 2 dimensional image in it's Albanese.  I have heard Catanese ask about the case where S is an abelian surface.  
A: My instinct tells me no. Here's a pseudo-proof. Take $X$ to be a surface with nontrivial
fundamental group e.g.
a product of two curves $X=C_1\times C_2$
where say $C_2$ has positive genus. Suppose it's birational to surface $Y\subset \mathbb{P}^3$
with isolated singularities at $p_1,\ldots p_N$. Let $U=\mathbb{P}^3-\{p_1,\ldots p_N\}$.
Then $\pi_1(U)=\pi_1(\mathbb{P}^3)$ is trivial. Then some version of Zariski-Lefschetz
should give $\pi_1(S\cap U)=\pi_1(U)=1$ (this is the "pseudo" part, since I'm too lazy to
track this down). We have a diagram
$$U\leftarrow U''\to U'\subset X$$
where the arrows are blow ups of points, and the inclusion is as an open set.
Then 
$$\pi_1(U)\cong \pi_1(U'')\cong \pi_1(U')$$
is trivial. However $\pi_1(U')$ would surject onto $\pi_1(X)$ QED.
Remark: It suffices to use $H_1$ in the place of $\pi_1$. I think this would be easier to
justify.
Remark 2: As is clear from remarks below, this argument is insufficient even with $H_1$, but perhaps there is a germ of a correct idea.
Some hours later: I no longer feel  that this approach is viable. Nevertheless,
I believe  for whatever irrational reason that there must be a counterexample.
One thing is certainly clear, and that is that this is a damn good problem.
A: This answer is completely rewritten. This is not an actual answer but 
a thought related to the question. I decided to leave it hear since it is short.
Note first that if there is a regular map from a surface $X$ to $\mathbb P^3$
whose image has only isolated singularities, then $X$ has curves with 
negative self-intersection. In particular, if $X$ has no such curves
then its image in $\mathbb P^3$ is smooth.
Now, suppose we have a surface $X$ with isolated singularities in $\mathbb CP^3$,
say of general type and consider the question:
Question. Let $X'$ be the minimal resolution of singularities on $X$.
Can we say something about $X$ if $X'$ contains rational $-1$ curves?
A: Let $S' \subset \mathbb{P}^3$ be the birational projection of a smooth surface $S \subset \mathbb{P}^4$. The general projection theorem of Gruson-Peskine (http://arxiv.org/abs/1010.2399v1) tells you that $S'$ is either smooth or has a curve of double points.
For instance if $S$ is the Severi surface in $\mathbb{P}^4$, then its projection on $\mathbb{P}^3$ (the Steiner surface) has a curve of double points.
So the answer to your question is "never true", unless your surface naturally lives in $\mathbb{P}^3$.
Edit Ok after some times, I realize that you are interested in a smooth surface (say $S \subset \mathbb{P}^5$) which maps birationnaly to a surface $S' \subset \mathbb{P}^3$ but for which the map does not come from the ambiant projective space. So my answer is useless...
