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 1. For a field $k$, every $k$-subalgebra $R$ of $k[x]$ that strictly contains $k$ is a finite type $k$-algebra, and $k[x]$ is a finitely generated $R$-module. Thus, every nonconstant, dominant $k$-morphism to an integral affine $k$-scheme from $\mathbb{A}^1_k$ is finite, hence surjective.
Proof. This is a variant of the proof of Noether's bound in invariant theory. Let $m(x)\in R$ be any nonconstant monic element. Then the $k$-subalgebra $S$ of $R$ generated by $m(x)$ is a finite type $k$-algebra. Moreover, the element $x\in k[x]$ satisfies a monic polynomial in $t$ with coefficients in $S$, namely $m(t) - m(x)$.
Thus $k[x]$ is a finitely generated $S$-module, and $R$ is an $S$-submodule. Since $S$ is a finite type $k$-algebra, hence Noetherian by the Hilbert Basis Theorem, the $S$-submodule $R$ is finitely generated. Hence $R$ is a finite type $k$-algebra, and the finitely generated $S$-module $k[x]$ is also finitely generated as an $R$-module. QED
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
Denote by $D$ the closed complement $X\setminus U$ with its reduced structure.
Lemma 3. The morphism $p$ is not finite if and only if $D$ is a nonempty divisor in $X$.
Proof. Of course the statement that $p$ is finite is equivalent to the statement that $D$ is empty.
Since $X$ is normal and affine over $Y$, the (reduced) closed complement $D=X\setminus U$ of the $Y$-affine scheme $U$ is either empty or has pure codimension $1$. 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.