Here is another way to do it.
By the Gödel-Rosser theorem, there are continuum many distinct consistent completions of PA. One can see this by building a tree of finite extensions of PA, using the Gödel-Rosser theorem at each node to branch with the Rosser sentence or its negation relative to that theory (and also deciding the $n^{\rm th}$ sentence), so that every branch through the tree is a complete consistent extension of PA. Every such consistent completion of PA has a countable model. Since different complete theories cannot have isomorphic models, you get continuum many non-isomorphic countable model of PA.
(Meanwhile, Andreas's answer applies not just to PA, but to any fixed theory, and so in fact, the compactness argument he mentions shows that each of these continuum many extensions of PA has continuum many non-isomorphic models.)
The answer to the question is in the positive.
Let $(\cal{N}^*,\textrm{Tr*})$ be a rather classless elementary extension of $(\cal{N},\textrm{Tr})$, where $\cal{N}$ is a model of PA, and $\textrm{Tr}$ is a full truth predicate on $\cal{N}$, e.g., let $\cal{N}$ be the standard model of PA, and $\textrm{Tr}$ be the usual Tarskian truth predicate on $\cal{N}$.
In the above, "rather classless" means that $(\cal{N}^*,\textrm{Tr*})$ has the property that every subset $X$ of the universe $N^*$ of $\cal{N}^*$ that is piecewise coded in $\cal{N}^*$ is first order definable in $(\cal{N}^*,\textrm{Tr*})$, where $X$ is piecewise coded means that the intersection of $X$ with every topped initial segment of $\cal{N^*}$ is parametrically definable in $\cal{N^*}$ (indeed, it is coded by a single element).
See Theorem 2.2.14 (and the preceding two paragraphs) of the Kossak-Schmerl monograph on models of PA for the relevant existence theorem for rather classless models. Such models exist in every uncountable cardinality $\kappa$ of uncountable cofinality and indeed can be arranged to be $\kappa$-like.
On the other hand, it is easy to see that every inductive set is piecewise coded. Therefore, if $\textrm{T}$ is an inductive truth predicate on $N^*$, then $\textrm{T}$ is piecewise coded and by rather classlessness it is definable in $(\cal{N}^*,\textrm{Tr*})$, and in particular the double-expansion $(\cal{N}^*,\textrm{Tr*}, \textrm{T})$ satisfies full induction.
Now, an easy induction carried out internally in $(\cal{N}^*,\textrm{Tr*}, \textrm{T})$ on the complexity of formulae shows that $\textrm{Tr*} = \textrm{T}$. QED
Best Answer
To move this off the unanswered queue, the answer is yes in a very strong way.
The relevant notion is that of a PA degree. There are many equivalent definitions of PA-ness, as well as many interesting results about them, but for the current purposes the relevant ones are the following I think:
Definition: d is PA iff every infinite computable subtree of $2^\mathbb{N}$ has a d-computable infinite path.
Theorem (Shoenfield?): The following are equivalent for each degree d:
d is PA.
There is a d-computable nonstandard model of PA (this justifies the name, and explains the relevance to the current question).
Theorem: there are surprisingly simple PA-degrees - namely ones $\le_T\emptyset'$ (Shoenfield), $<_T\emptyset'$ (Kreisel), and even low (Jockusch/Soare). And there are other basis theorems too, but these three in order tell the simplest story in my opinion.
Consequently, any modification of the Turing machine model in which the usual halting problem is newly computable will be able to construct nonstandard models of PA. ITTMs, OTMs, and every other variation on the basic theme of infinitary computation goes well beyond this; the closest variant I can think of is limit computability, which Shoenfield showed corresponds to computability relative to the halting problem.
See e.g. Diamondstone/Dzhafarov/Soare for more information on PA degrees and the attendant notions.