Let $M = (N, \oplus, \otimes, <_M, 0_M, 1_M)$ be a nonstandard model of peano arithmetic. $\oplus$ and $\otimes$ are uncomputable due to Tennenbaum's theorem.
For $c \in N$, let $\oplus_{<c}, \otimes_{<c}$ be defined as follows: $a \oplus_{<c} b = a \oplus b$ if $a \oplus b <_M c$. Otherwise $a \oplus_{<c} b$ is undefined. $a \otimes_{<c} b = a \otimes b$ if $a \otimes b <_M c$. Otherwise $a \otimes_{<c} b$ is undefined.
Is there a nonstandard model $M$ of peano arithmetic and nonstandard $c \in N$ such that $\oplus_{<c}$ and $\otimes_{<c}$ are computable?
No.
Let $A, B$ be recursively inseparable r.e. sets. For any natural $k$, it is true that
$$M \models \exists y. y < c \land \forall i<x . (i \in A \implies p_i \mid y) \land (i \in B \implies p_i \nmid y)$$
(where $p_i$ is the $i$th prime) so by overspill, there is a nonnatural $k$ with the above property. Let $y$ be the witness of the associated statement.
Now, for any natural $i$, we can calculate $p_i$, and if $\oplus_{<c}$ is computable, we can look for $q, r$ such that:
$(\underbrace{q \oplus q \oplus \dots \oplus q}_{p_i\text{ times}})\oplus r = y <_M c$.
If $r=0$, $r \in C$. Otherwise, $r \notin C$. $C$ recursively separates $A$ and $B$. Contradiction. $\square$