If perturbations are sufficiently small and $\rho_{AB}$ has sufficiently broad support then a desired global state $\rho_{AB}'$ exists. Define
$$
\rho_{AB}' = \rho_{AB} + (\rho_A' - \rho_A) \otimes \rho_B + \rho_A \otimes (\rho_B' - \rho_B) \tag1.
$$
Note that $\rho_{AB}'$ is Hermitian and trace one, but may not be positive. However, $\rho_{AB}'$ is positive if $\rho_{AB}$ has broad support and perturbations are not too large. For example, if
$$
\|\rho_A' - \rho_A\|_2 + \|\rho_B' - \rho_B\|_2 \leq \lambda_{min}(\rho_{AB})
$$
where $\lambda_{min}(X)$ denotes the smallest eigenvalue of operator $X$, then for any $|\psi\rangle$
$$
\begin{align}
\langle\psi|\rho_{AB}'|\psi\rangle
& =
\langle\psi|\rho_{AB}|\psi\rangle +
\langle\psi|(\rho_A' - \rho_A) \otimes \rho_B|\psi\rangle +
\langle\psi|\rho_A \otimes (\rho_B' - \rho_B)|\psi\rangle \\
& \geq
\lambda_{min}(\rho_{AB}) - \|\rho_A' - \rho_A\|_2 - \|\rho_B' - \rho_B\|_2 \geq 0.
\end{align}
$$
If $\lambda_{min}(\rho_{AB}') = 0$, then it may be possible to restrict the support of the perturbations so that an analogous inequality holds.
The reduced states of $\rho_{AB}'$ are
$$
\begin{align}
{\rm tr}_A(\rho_{AB}')
& = {\rm tr}_A(\rho_{AB}) + \rho_B \, {\rm tr}(\rho_A' - \rho_A) + (\rho_B' - \rho_B) \, {\rm tr}(\rho_A) \\
& = \rho_B + \rho_B' - \rho_B \\
& = \rho_B'
\end{align}
$$
and similarly ${\rm tr}_B(\rho_{AB}') = \rho_A'$.
Finally, the distance
$$
\begin{align}
\|\rho_{AB}' - \rho_{AB}\|_1
& = \|(\rho_A' - \rho_A) \otimes \rho_B + \rho_A \otimes (\rho_B' - \rho_B)\|_1 \\
& \leq \|(\rho_A' - \rho_A) \otimes \rho_B\|_1 + \| \rho_A \otimes (\rho_B' - \rho_B)\|_1 \\
& = \|\rho_A' - \rho_A\|_1 + \|\rho_B' - \rho_B\|_1 \\
& \leq 2\delta
\end{align}
$$
and
$$
\lim_{\delta\rightarrow 0}\varepsilon(\delta) = \lim_{\delta\rightarrow 0} 2\delta = 0
$$
as required.
Note that the above construction fails for very pure highly entangled states since in this case the reduced states are nearly completely mixed and so the two product terms in $(1)$ will contain negative elements that are not compensated for by $\rho_{AB}$.