Other answers showed that Cantor–Bernstein–Schroeder theorem is not valid constructively in the following sense: there exist models of Intuitionistic Zermelo–Fraenkel set theory for which Cantor–Bernstein–Schroeder property does not hold.
Of course, this does not mean that Cantor–Bernstein–Schroeder property is incompatible with constructive mathematics --- it just shows that IZF is too weak to prove CBS. Therefore, a complementary question could be: what are the implications of IZF+CBS; does IZF+CBS make the logic collapse to the boolean case?
If I am not mistaken, the answer is no, and the counterexample is constructed below. However, I shall start with a negative observation.
Let $\Omega$ be a Heyting algebra with countable unions. An element $v \in \Omega$ is complementable if there exists an element $w \in \Omega$ such that $v \vee w = 1$ and $v \wedge w = 0$.
We say that $\Omega$ is a boolean algebra if its every element is a finite union of complementable elements (equivalently, if every element is complementable).
We say that $\Omega$ is a pro-boolean algebra if its every element is a countable union of complementable elements.
The claim is that every elementary topos with countable colimits satisfying Cantor–Bernstein–Schroeder propery is pro-boolean (i.e. its subobject classifier is a pro-boolean algebra).
Let $v \colon V \rightarrow 1$ be a ``truth'' value (a monomorphism into terminal object). We shall construct two objects $X = \coprod_{\mathbb{N}} 1$ and $Y = V \sqcup X$. There are obvious monomorphisms ${\iota_X}\colon{X}\rightarrow{Y}$ given by the coproduct injection, and ${v \sqcup \mathit{id}}\colon{Y}\rightarrow{X}$. Therefore, by Cantor–Bernstein–Schroeder there is an isomorphism ${b}\colon{Y}\rightarrow{X}$. Since coproducts in a topos are extensive, we may divide each component ${\iota_0}\colon V \rightarrow{Y}$, ${\iota_k}\colon 1 \rightarrow{Y}$ of the coproduct $Y$ along $b$ through coproduct's injections ${\iota_l} \colon 1 \rightarrow{X}$ obtaining elements $\alpha_{k,l}$ such that $Y \approx \coprod_k \coprod_l \alpha_{k,l} \approx \coprod_l \coprod_k \alpha_{k,l}$ and $b = \coprod_l b_l$ with each ${b_l}\colon{\coprod_l \alpha_{l,k}}\rightarrow{1}$ being an isomorphism (using again extensivity of coproducts and the fact that pullback of an iso is iso). Because unions in a topos are effective (and coproducts are disjoint) $1 \approx \coprod_l \alpha_{l,k} = \bigcup_l \alpha_{l,k}$ and so each $\alpha_{l,k}$ is complemented by $\bigcup_{x \neq l} \alpha_{x,k}$. Since every subterminal value can sit in exactly one way in the terminal object $v = \bigcup_l \alpha_{l,0}$ is a countable disjoint union of complementable elements.
(we cannot get more from this construction; for example in the category of sheaves over rational numbers with the usual topology, $\coprod_{\mathbb{N}} 1 \approx v \sqcup \coprod_{\mathbb{N}} 1$ for (non-complementable) truth value $v$ corresponding to the open ball $(-1, 1)$; however there are other objects in this category that can serve as counterexamples for CBS)
For a positive result, consider set: $$\mathcal{D} = \lbrace 0, 1, \frac12, \frac13, \frac14, \dotsc\rbrace$$ with topology inherited from $\mathbb{R}$ and construct the category $\mathit{Sh}(\mathcal{D})$ of sheaves over $\mathcal{D}$. Every open set in $\mathcal{D}$ can be build from singletons $\lbrace\frac1n\rbrace$ and a set of the form $[0, \frac1k]$.
Let $F, G \colon \mathcal{D}^{op} \rightarrow \mathbf{Set}$ be any sheaves and assume there are monomorphisms $m \colon F \rightarrow G$ and $n \colon G \rightarrow F$. A monomorphism between sheaves is an injection on its each component, therefore by CBS theorem for sets $F(U) \approx G(U)$ for every open set $U$. Let $\phi_U \colon F(U) \approx G(U)$ be a collection of such isomorphisms. We shall construct an isomorphism $\alpha \colon F \rightarrow G$ between sheaves inductively:
$\lambda_{[0, 1]} = \phi_{[0, 1]}$
for every nonempty $F(\lbrace\frac1k\rbrace)$ choose an element $1_{\frac1k} \in F(\lbrace\frac1k\rbrace)$; if $F(\lbrace\frac1k\rbrace)$ is empty then $\lambda_{[0, \frac1{k+1}]} = \phi_{[0, \frac1{k+1}]}$, otherwise $\lambda_{[0, \frac1{k+1}]} = G([0, \frac1{k+1}] \subset [0, \frac1{k}]) \circ h_{\frac1k}$, where $h_{\frac1k} \colon F([0, \frac1{k+1}]) \rightarrow F([0, \frac1{k}])$ is the unique morphism to the product $F([0, \frac1{k}]) = F([0, \frac1{k+1}]) \times F(\lbrace\frac1k\rbrace)$ induced by $F([0, \frac1{k+1}]) \overset{!}\rightarrow 1 \overset{1_{\frac1k}}\rightarrow F(\lbrace\frac1k\rbrace)$ and the identity on $F([0, \frac1{k+1}])$
similarly, for every nonempty $F([0, \frac1{k+1}])$ choose an element $1_{[0, \frac1{k+1}]} \in F([0, \frac1{k+1}])$; if $F([0, \frac1{k+1}])$ is empty then $\lambda_{\lbrace\frac1{k}\rbrace} = \phi_{\lbrace\frac1{k}\rbrace}$, otherwise $\lambda_{\lbrace\frac1{k}\rbrace} = G(\lbrace\frac1{k}\rbrace \subset [0, \frac1{k}]) \circ h_{1_{[0, \frac1{k+1}]}}$
if $U$ is a disjoint union of the form $[0, \frac1k] \sqcup \bigcup_i\lbrace\frac{1}{n_i}\rbrace$, where $[0, \frac1k]$ is the largest interval contained in $U$, then $\lambda_{U} = \lambda_{[0, \frac1k]} \times \prod_i \lambda_{\lbrace\frac{1}{n_i}\rbrace}$, where the products are determined by structures of the sheaves.
In the second and third step we have chosen the components of $\lambda$ to be upward compatible, and in the fourth step the naturality condition follows from the universal property of products. Thus $F \approx G$.
(BTW, I think we are not really that far from the inverse of the above theorem, but that is for another story...)