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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.

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.

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.

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.

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.

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.