Thanks to Omar's example illustrating the salient point, here's a proof of Bredon's statement.
Let $\mathcal C = \mathcal B \cup (X \times \{0\})$ as a point set. A topology on $\mathcal C$ is generated by the open sets of $\mathcal B$ together with all sets of the form $U\times\{0\}$ for $U\subseteq X$ open. Writing $\pi_\mathcal B:\mathcal B\to A$ and $\pi_0:X\times\{0\}\to X$ for the local homeomorphisms of the sheaves $\mathcal B$ and $X\times\{0\}$ respectively, we wish to define a local homeomorphism $\pi:\mathcal C\to X$. There are no problems with continuity of the group operations, so such a map will make $\mathcal C$ a sheaf.
Let of course \begin{align} \pi(c) =\begin{cases} \pi_{\mathcal{B}}(c) \quad &\text{if } c\in \mathcal{B}\\ \pi_0(c) \quad &\text{if } c\in X\times\{0\}. \end{cases} \end{align}
Let $c\in\mathcal C$. If $c\notin\mathcal B$, then everything is OK, so assume $c\in\mathcal B$. Write $C$ for the neighborhood (in $\mathcal B$) on which $\pi_\mathcal{B}$ is a homeomorphism. Let $D$ be an $X$-open set such that $\pi_\mathcal{B}(C)=D\cap A$.
If we assume that $A$ is locally closed, then there is a $X$-open neighborhood $U$ of $\pi(c)\in A$ such that $U\cap(X\setminus A)$ is open in $U$. Thus so is $U\cap D\cap (X\setminus A)$, and we can write $U\cap D\cap(X\setminus A)=U\cap D\cap V$ for some $X$-open set $V\subseteq X\setminus A$. Then $\pi_0^{-1} (V) = V\times\{0\}$ is open in $\mathcal C$, and $\pi$ is a homeomorphism on the open neighborhood $(\pi_\mathcal{B})|_C^{-1}(U\cap D\cap A)\cup\pi_0^{-1}(V)$ of $c$.