Return to Answer

By hypothesis on $E$, the element $p^*f$ is irreducible in $k[x_1,\dots,x_n]$. Thus, the invertible elements of $k[x_1,\dots,x_n][p^*f^{-1}]$ are precisely the elements of the form $cf^d$ for $c\in k^\times$ and for $d\in \ZZ$. Since $g$ is invertible in $k[x_1,\dots,x_n][p^*f^{-1}]$, the element $g$ equals $cf^d$ for some $c$ and integer $d$. Thus, $g$ is already invertible in $k[X][\overline{p}^*f^{-1}]$. This contradicts the hypothesis that $E$ does not contain $\overline{p}(D)$. QED

By hypothesis on $E$, the element $p^*f$ is irreducible in $k[x_1,\dots,x_n]$. Thus, the invertible elements of $k[x_1,\dots,x_n][p^*f^{-1}]$ are precisely the elements of the form $cf^d$ for $c\in k^\times$ and for $d\in \mathbb{Z}$. Since $g$ is invertible in $k[x_1,\dots,x_n][p^*f^{-1}]$, the element $g$ equals $cf^d$ for some $c$ and integer $d$. Thus, $g$ is already invertible in $k[X][\overline{p}^*f^{-1}]$. This contradicts the hypothesis that $E$ does not contain $\overline{p}(D)$. QED

The Ax-Grothendieck Theorem is far more general than for the case of a morphism from affine space to itself. Here is a sketch of the proof of the result in that case. The basic ideas are similar to those in Jouanolou's Bertini-theory book used to simplify the proof of the Quillen-Suslin Theorem.

Edit. I cannot complete my proof. The result is wrong, as shown by the accepted answer by user @pinaki. The problem with my proof is that the divisor $\text{Zero}(f)$ I need in the target must simultaneously be transverse to the branch divisor and contain the image of the singular locus of the integral closure variety $X$. Those two conditions are mutually exclusive. I am keeping the post below because it proves the result for $n=1$, and it highlights the issue for $n>2$.

Lemma 2. For a field $k$, for every affine, dominant, quasi-finite $k$-morphism $p:U\to Y$ between normal, integral, finite type $k$-schemes, there is a factorization of $p$ as $i:U\to X$ composed with $\overline{p}:X\to Y$, where $i$ is a dense open immersion, and $\overline{p}$ is a finite $k$-morphism between normal, integral, finite type $k$-schemes.

Proof. Of course this follows quickly from Grothendieck's version of Zariski's Main Theorem. It also follows from the Noether Normalization Theorem: let $X$ be the integral closure of $Y$ in the function field of $U$. QED

Hypothesis 4. The non-smooth locus of each $k$-scheme $U$ and $Y$ has irreducible components of codimension $\geq 2$.

Since $U$ and $Y$ are normal, this follows from Serre's Criterion for Normality if the characteristic is $0$. Of course in the case of interest, when $U$ and $Y$ are each isomorphic to affine space, the hypothesis holds.

By construction of $X$, also the non-smooth locus has codimension $\geq 2$. Thus, there is a closed subscheme $Z\subset Y$ all of whose irreducible components have codimension $\geq 2$ such that $Y\setminus Z$, and the preimage $\overline{p}^{\text{pre}}(Y\setminus Z)$ are smooth $k$-schemes, and the intersection of these smooth $k$-schemes with $D$ is reduced (possibly reducible). In particular, if $D$ is not empty, then $D$ is a Cartier divisor on the open $\overline{p}^{\text{pre}}(Y\setminus Z)$. Since $X$ is finite over $Y$, this Cartier divisor becomes principal, say the divisor of $g$, on all preimage open subsets $\overline{p}^{\text{pre}}(V)$ for $V$ sufficiently small Zariski open subsets in $Y\setminus Z$. In particular, if $V$ equals $D(f)$, then the ring extension $k[Y][f^{-1}] \to k[U][f^{-1}]$ contains $k[X][f^{-1},g^{-1}]$ as a subring, i.e., $g$ is already invertible in $k[U][f^{-1}]$. However, when $k[U]$ equals $k[x_1,\dots,x_n]$, we know what are all the invertible elements in $k[U][f^{-1}]$.

The argument from here on uses a Bertini result. I am trying to write this up now.

Edit. The Bertini result that I needed is wrong; see the explanation above.

Edit. To try and make something productive out of the argument below, let me explain the proof of the Ax-Grothendieck Theorem for affine space using this argument. Also, the argument below includes the proof of the conjecture of the OP when $n$ equals $1$. As noted in the accepted answer, the conjecture is false when $n$ equals $2$. The argument below shows that when the result fails, the image of the morphism is an open subset of affine space whose closed complement has codimension $\geq 2$, containing the image of the singular locus of the "integral closure" variety $X$ and intersecting the singular locus of the branch divisor (this is why the Bertini result fails).

The proof strategy here is similar to that used by Jouanolou in his book on Bertini theorems to simplify the proof of the Quillen-Suslin Theorem.

Lemma 2. For a field $k$, for every affine, dominant, quasi-finite $k$-morphism $p:U\to Y$ between normal, integral, finite type $k$-schemes, there is a factorization of $p$ as $i:U\to X$ composed with $\overline{p}:X\to Y$, where $i$ is a dense open immersion, and where $\overline{p}$ is a finite $k$-morphism between normal, integral, finite type $k$-schemes.

Proof. This follows from Grothendieck's version of Zariski's Main Theorem and the Noether Normalization Theorem: $X$ is the integral closure of $Y$ in the function field of $U$, this is finite type over $Y$, and $U$ is an open subscheme of $X$. QED

Hypothesis 4. The field $k$ is perfect, both $U$ and $Y$ are smooth $k$-schemes, and $p$ is generically étale.

By construction of $X$, also the non-smooth locus $X_{\text{sing}}$ of $X$ has codimension $\geq 2$. On the smooth open complement $X^o:=X\setminus X^{\text{sing}}$, the reduced Weil divisor is Cartier, i.e., the ideal sheaf is locally principal. Since $\overline{p}$, the ideal sheaf is even principal when restricted to $\overline{p}$-preimages of sufficiently small Zariski open subsets of $Y$, e.g., the inverse image $\overline{p}^{\text{pre}}(Y\setminus E)$ for a divisor $E$ in $Y$ that contains the image $\overline{p}(X_{\text{sing}})$. If $D$ is nonempty, then we can choose $E$ so that it does not contain $\overline{p}(D)$, i.e., so that the restriction of $D$ to the preimage open subset is nonempty and the generator of the principal ideal is a nonzero nonunit.

Problem 5. Assuming $D$ is nonempty, among irreducible divisors $E$ in $Y$ containing $\overline{p}(X_{\text{sing}})$ yet not contain $\overline{p}(D)$, and such that $D$ is principal on $\overline{p}^{\text{pre}}(Y\setminus E)$, is there one such that $\overline{p}^{\text{pre}}(E)$ is irreducible?

Lemma 6. If $p$ is birational, then there is such a divisor $E$.

Proof. If $p$ is birational, then by Zariski's Main Theorem, also $\overline{p}$ is an isomorphism. In particular, it is a homeomorphism for the Zariski topology, and inverse images of irreducible closed subsets are irreducible. QED

Proposition 7. For a dominant,quasi-finite $k$-morphism $p:U\to Y$ between affine spaces over $k$, if there exists a divisor $E$ as in the problem, then already $p$ is finite, hence surjective. In particular, if $p$ is birational, then $p$ is an isomorphism of $k$-schemes.

Proof. The morphism $p$ is finite if and only if the divisor $D$ is nonempty. By way of contradiction, assume that $D$ is nonempty and a divisor $E$ as in the problem exists. Since $Y$ is factorial (by Gauss's Lemma), the irreducible divisor $E$ is $\text{Zero}(f)$ for some irreducible $f\in k[Y] \cong k[y_1,\dots,y_n]$. By hypothesis, in the $k$-algebra $k[X][f^{-1}]$, the ideal of $D$ is principal, say $\langle g\rangle$. Thus, the fraction ring $k[X][\overline{p}^*f^{-1},g^{-1}]$ is already contained in the $k$-subalgebra $k[U][p^*f^{-1}] \cong k[x_1,\dots,x_n][p^*f^{-1}]$.

By hypothesis on $E$, the element $p^*f$ is irreducible in $k[x_1,\dots,x_n]$. Thus, the invertible elements of $k[x_1,\dots,x_n][p^*f^{-1}]$ are precisely the elements of the form $cf^d$ for $c\in k^\times$ and for $d\in \ZZ$. Since $g$ is invertible in $k[x_1,\dots,x_n][p^*f^{-1}]$, the element $g$ equals $cf^d$ for some $c$ and integer $d$. Thus, $g$ is already invertible in $k[X][\overline{p}^*f^{-1}]$. This contradicts the hypothesis that $E$ does not contain $\overline{p}(D)$. QED

Nota bene. In my original post, I attempted to use Bertini theorems to find $E$ as above: the inverse image of $E$ is connected by the Bertini connectedness theorem. If $E$ intersects transversally the branch divisor of $\overline{p}$ at all generic points of the intersection set, then the inverse image of $E$ is also locally irreducible, and hence $E$ is irreducible. However, as the example in the accepted answer shows, it can happen that $\overline{p}(X_{\text{sing}})$ contains an irreducible component of the singular locus of the branch divisor that has codimension $2$ in $Y$. Thus, the requirement that $E$ contains $\overline{p}(X_{\text{sing}})$ forces $E$ to intersect the branch divisor nontransverally at the generic point of $\overline{p}(X_{\text{sing}})$.

The Ax-Grothendieck Theorem is far more general than for the case of a morphism from affine space to itself. Here is a sketch of the proof of the result in that case. The basic ideas are similar to those in Jouanolou's Bertini-theory book used to simplify the proof of the Quillen-Suslin Theorem. I discussed some part of this proof many years ago with Alan Durfee. I believe that we found a reference at that time; I will try to dig it up. Edit. I am trying to write a careful proof of the Bertini theorem I used in the original post. Until then, here is the proof for $n=1$ and the setup for $n\geq 2$.

Edit. The argument from here on uses a Bertini result. I am trying to write this up now.

The Ax-Grothendieck Theorem is far more general than for the case of a morphism from affine space to itself. Here is a sketch of the proof of the result in that case. The basic ideas are similar to those in Jouanolou's Bertini-theory book used to simplify the proof of the Quillen-Suslin Theorem. 

Edit. I cannot complete my proof.  The result is wrong, as shown by the accepted answer by user @pinaki. The problem with my proof is that the divisor $\text{Zero}(f)$ I need in the target must simultaneously be transverse to the branch divisor and contain the image of the singular locus of the integral closure variety $X$. Those two conditions are mutually exclusive. I am keeping the post below because it proves the result for $n=1$, and it highlights the issue for $n>2$.

The argument from here on uses a Bertini result. I am trying to write this up now.

Edit. The Bertini result that I needed is wrong; see the explanation above.