Suppose we want to prove CSB theorem in $\mathsf{KP}$ following this answer. Let $f\colon A \to B$ and $g\colon B \to A$ be injections. Let $C_0 = A \setminus g(B)$ and $C_{n + 1} = g(f(C_n))$.
The $C_n$s are defined in the answer as $$C_n = \{x \in A : \exists s(s:\{0,\ldots,n\} \to A \land s(0) \in C_0 \land (\forall i < n)(s(i+1)=g(f(s(i)))) \land s(n)=x\}$$ I think $$C_n = \{x \in A : \forall s(\operatorname{dom}(s) = n + 1 \land s(0) = C_0 \land (\forall i < n)(s(i+1)=g[f[s(i)]]) \land x \in s(n)\}$$ also works thus $C_n$ are uniformly defined by a $\Delta_1$ formula.
Edit: Now I think this isn't trivial. How could one show that such $s$ exists for all $n$? Stating that such $s$ exists is would be a $\Sigma_1$ formula but $\mathsf{KP}$ doesn't have $\Sigma_1$-separation.
Now let's define a class $C'$ following the answer as $$C' = \{x \in A: \forall D, (C_0 \subseteq D \subset A \land g[f[D]] \subset D \to x \in D)\}$$ Since $C_n$s are defined by a $\Delta_1$ formula, we can use mathematical induction on $n$ to prove that $\forall n$ $C_n \subset C'$. On the other hand, $\bigcup_n C_n$ satisfies the above condition for $D$, so $C' \subset \bigcup_n C_n$. Thus $\bigcup_n C_n = C'$, and they are really sets by $\Delta_1$-separation. Then
$$h(x) = \begin{cases} f(x) & x \in C \\ g^{-1}(x) & x \in A \setminus C \end{cases}$$
is the desired bijection between $A$ and $B$. Is this correct?
For a technical convenience, let us assume that $B\subseteq A$ and $g$ is the insertion map. We can make such an assumption even over $\mathsf{KP}$.
Now take $C_0=A\setminus B$, $C_{n+1}=f[C_n]$, which can be defined by $\Sigma$-recursion. The sequence $\langle C_n\mid n<\omega\rangle$ is $\Sigma$-definable, so it is a set by $\Sigma$-Replacement. That means we can define the set $C=\bigcap_{n<\omega} C_n$. Now we can carry over the usual construction so that we have a bijection $h\colon A\to B$ by setting $$h(x) = \begin{cases} f(x) & \text{if }x\in A\setminus C, \\ x & \text{if }x\in C.\end{cases}$$
You can see that the above argument works over $\mathsf{KP}$ because what I used are $\Sigma$-recursion and $\Sigma$-Replacement, which are provable over $\mathsf{KP}$.
However, Cantor-Bernstein theorem is less useful over $\mathsf{KP}$ because $\mathsf{KP}$ is compatible with the countability of every set. $H_{\omega_1}$ is one model for $\mathsf{KP}$ + "Every set is countable." (In fact, $H_{\omega_1}$ satisfies $\mathsf{ZFC}^-$, $\mathsf{ZFC}$ without Powerset but with Collection.)
Furthermore, $\mathsf{KP}$ cannot make use of definable injections between sets that are not provably sets. The above proof assumes $f$ is a set function, and the above argument works even when $f$ is $\Sigma$-definable class function. However, $\Sigma$-Replacement proves any $\Sigma$-definable class functions between sets are sets.