First, let's talk about what Abramsky means when he says that measurements are "locally consistent". Suppose we have a set of three measurements $M_{1},M_{2},M_{3}$ every which can take only values $\pm 1$, and our contexts are given by $\{ \, \{M_{1},M_{2} \}, ,\ \{M_{2},M_{3}\} \, \}$, that is, only $M_{1}$ and $M_{3}$ are not compatible. If we set the behavior of our system(the context's probabilities) to be the extreme distributions $p_{M_{1},M_{2}}(+1,+1) = 1$ and $p_{M_{2},M_{3}}(-1,-1) = 1$ we get
different probabilities for $M_{2}$ when we marginalize $p_{M_{1},M_{2}}$ and $p_{M_{2},M_{3}}$. We say that this is a disturbing scenario and this is what Abramsky would call "locally inconsistent". It happens that many theories, in particular classical probability theory and quantum mechanics, are non-disturbing. That is, if the measurement $M$ is in the context $C_{1}$ as well as in context $C_{2}$, we have
$$
Prob(M = m | C_{1}) = Prob(M = m |C_{2})
$$
Where $Prob(M=m|C)$ is just the probability of finding $M=m$ when we marginalize though the context $C$. Or, in general, if $U \subseteq C_{1} \cap C_{2}$, $p_{U}^{C_{1}} = p_{U}^{C_{2}}$ where $p_{U}^{C}$ is the probability distribution for finding a specific set of value for the measurements in $U$ obtained by marginalization in $C$.
If we accept only non-disturbing models and we suppose that every measurement is compatible, we will have only one big maximal context $C$ such that every measurement is in it. Therefore, of course, we can define a global distribution of probabilities that is simply $p_{C}$ and the distribution of every set of measurements is defined $p_{U} = p_{U}^{C}$. We have no contextuality effects!
This is enough to show that in quantum mechanics the answer is yes, we do need incompatibility for contextuality effects to arise. If, on the other hand, we allow for disturbing models, we are left with the question: if I have a set of compatible measurements $C$ and $U \subseteq C$, can I say that $p_{U}^{C} = p_{U}$? The answer is yes.
Proof:
Let's assume that our context is finite $C = \{ M_{1},..., M_{N} \}$ and that, without loss of generality, $U = \{ M_{1},...,M_{u}\}$ with $u\leq N$. Since $C$ is a context, it has a refinement $M$ (that is, a measurement) associated with it along with functions $f_{1},f_{2},...,f_{N}$ such that
$$
Prob(M_{1} = m_{1}, ... , M_{N} = m_{N}|C) = \sum_{f_{1}(m) = m_{1} \, and \, \cdots \, and f_{N}(m)=m_{N}} Prob(M = m)
$$
and, as well,
$$
Prob(M_{1} = m_{1}, ... , M_{u} = m_{u}|C) = \sum_{f_{1}(m) = m_{1} \, and \, \cdots \, and f_{u}(m)=m_{u}} Prob(M = m)
$$
what we wish to prove is that
$$
Prob(M_{1} = m_{1}, ... , M_{u} = m_{u}|C) = \sum_{m_{u+1},\cdots,m_{N}} Prob(M_{1} = m_{1}, ... , M_{N} = m_{N}|C)
$$
Well,
\begin{align*}
\sum_{m_{u+1},\cdots,m_{N}} Prob(M_{1} = m_{1}, ... , M_{N} = m_{N}|C) &= \sum_{m_{u+1},\cdots,m_{N}} \left( \sum_{f_{1}(m) = m_{1} \, and \, \cdots \, and f_{N}(m)=m_{N}} Prob(M = m) \right) \\
& \text{since we are adding every value of $m_{u+1},\cdots,m_{N}$} \\
&= \sum_{f_{1}(m) = m_{1} \, and \, \cdots \, and f_{u}(m)=m_{u}} Prob(M = m) \\
&= Prob(M_{1} = m_{1}, ... , M_{u} = m_{u}|C)
\end{align*}
The example of disturbing scenario and the formalism and notation used to define compatibility are taken from the book
Amaral, B., & Cunha, T. M. (2018). On Graph Approaches to Contextuality and their Role in Quantum Theory (SpringerBriefs in Mathematics) (1st ed. 2018 ed.). Springer. https://doi.org/10.1007/978-3-319-93827-1'
