Theorem 1.53 (3) in page 227 of Hajek and Pudlak's book, Metamathematics of First-Order Arithmetic, says:
Theorem. If $M$ is a countable model of $I\Delta_{0}$ such that $M$ has a proper elementary end extension, then $M\models PA$.
Is the above theorem still true if we drop the countability assumption of $M$?
The countability assumption cannot matter. The reason is that any uncountable model $M$ is countable in a forcing extension of the set-theoretic universe. If $M$ has a proper elementary end-extension in the original universe, then this end-extension still exists in the forcing extension. So we may apply the theorem as you have stated it in the forcing extension in order to deduce that $M\models\text{PA}$. But the question whether $M$ is a model of PA or not is absolute between the set-theoretic universe $V$ and its forcing extensions $V[G]$, and so $M\models\text{PA}$ in $V$, as desired.
Although I like this kind of use of forcing, where one constructs the forcing extension in order solely to make a conclusion about what is true in the original universe, nevertheless one may omit it by taking a countable elementary substructure of a suitable $H_\theta$, in effect replacing $M$ with a countable proxy. Applying the theorem to the proxy, you get that the proxy satisfies $\text{PA}$, and so $M\models\text{PA}$ as well.
The theorem remains true without the countability assumption. The quickest way to see this is to deduce it from the following well-known facts about the regularity scheme (see here for a definition of the regularity scheme).
1. Suppose $\cal{M }$ is an ordered structure (of any cardinality) that has a proper elementary end extension, then $\cal{M}$ satisfies the regularity scheme. This is a routine exercise.
2. A model $\cal{M}$ of $I\Delta_{0}$ (of any cardinality) satisfies $PA$ iff $M$ satisfies the regularity scheme. See, e.g., Theorem 7.3 of Kaye's text on models of $PA$ for a proof.
Remark. The converse of Lemma 1 is true for countable $\cal{M}$, and is due to Keisler. The proof uses the omitting types theorem. It is known that the converse fails without the countability assumption.
Having seen the other answers, let me point out yet another quick argument using just Lowenheim-Skolem.
Let $M$ be an uncountable model and $N$ a proper elementary end extension of it. Expand $N$ by adding an extra unary predicate naming the subset $M$. Call the resulting structure $(N,M)$. Take a countable elementary substructure of it, say $(N',M')$. Then $N'$ is a proper elementary end extension of $M'$ and $M'$ is countable, so the theorem applies. Hence $M'$ is a model of PA and so is $M$ since they are elementarily equivalent.
