*Yes*, it does.

The collection $I^{\omega_2}_{WD}= \{ S \subset \omega_2: \neg \Phi^{\omega_2}_S \}$ of subsets of $\omega_2$ is an ideal (a proof this fact is provided in the proposition below; moreover I think this result is originally due to Shelah, however a reference escapes me at the moment.)

Moreover, $\Phi_{\omega_2}$ fails, precisely when $I^{\omega_2}_{WD}$ is not a proper ideal, which in this case is equivalent to some club subset of $\omega_2$ being a member of $I^{\omega_2}_{WD}$.

So if $\Phi^{\omega_2}_S$ fails for both $E^{\omega_2}_\omega$ and $E^{\omega_2}_{\omega_1}$. Then $E^{\omega_2}_{\omega}\cup E^{\omega_2}_{\omega_1} =\lim(\omega_2) \in I^{\omega_2}_{WD}$ and this implies the failure of $\Phi_{\omega_2}$, which is equivalent to $2^{\aleph_1} = 2^{\aleph_2}$.

**Proposition:** For any cardinal $\lambda$, the set $I^{\lambda^{+}}_{WD} = \{S \subset \lambda^{+}: \neg \Phi^{\lambda^{+}}_S \}$ is a $\lambda^{+}$-complete ideal extending the non-stationary ideal on $\lambda^{+}$.

**Proof:** That $I_{WD}^{\lambda^{+}}$ contains the non-stationary ideal on $\lambda^{+}$ follows by definition. As such, all that remains is to show $I^{\lambda^{+}}_{WD}$ is an appropriately complete ideal.

To this end suppose $S \in I^{\lambda^{+}}_{WD}$ as witnessed by the sequence $\langle F_\alpha: \alpha \in S \rangle$ where for each $\alpha \in S$, $F_{\alpha}:\,^{\alpha}2\rightarrow 2$. Then, for every $g:\lambda^{+} \rightarrow 2$, there is some $f:\lambda^{+} \rightarrow 2$ and club $C_g \subset \lambda^{+}$ such that

$$ \{ \alpha \in S \cap C_g: F_\alpha(f\vert_\alpha) = g(\alpha) \} = \emptyset $$

as such, for every $S_0 \subset S$ we have $ \{ \alpha \in S_0 \cap C_g: F_\alpha(f\vert_\alpha) = g(\alpha) \} = \emptyset $, hence the sub-sequence $\langle F_\alpha : \alpha \in S_0\rangle$ witnesses $S_0 \in I^{\lambda^{+}}_{WD}$.

Next, fix a bijection $\varphi:\lambda^{+}\times \lambda^{+}\rightarrow \lambda^{+}$ and let $C = \{ \alpha \in \lim(\lambda^{+}): \varphi[\alpha \times \alpha] = \alpha \}$ be the club subset of $\lambda^{+}$ consisting of closure points of $\varphi$.
Now, assume $\{ S_\gamma: \gamma \in \delta \} \subset I_{WD}^{\lambda^{+}}$ (with $\delta \in \lambda^{+}$) are stationary, pairwise disjoint, and that for each $\gamma\in \delta$, the sequence $\langle F^{\gamma}_\alpha : \alpha \in S_\gamma \rangle$ witnesses the failure of $\Phi^{\lambda^{+}}_{S_\gamma}$.

Letting $E = \bigcup \{ S_\gamma : \gamma \in \delta \}$, define the sequence $\langle F^{\ast}_\alpha : \alpha \in E\rangle$ of functions $F^{\ast}_\alpha : \,^{\alpha}2\rightarrow 2$ as follows, given $\alpha \in E$ and $h \in \,^{\alpha}2$:

- if $\alpha \in C\cap S_\gamma$, set $F^{\ast}_{\alpha}(h) = F^{\gamma}_{\alpha}([h]_\gamma)$, where $[h]_\gamma$ denotes the function defined by $[h]_\gamma(\xi) = h(\varphi(\gamma,\xi))$, and if $\alpha \not\in C$, simply take $F^{\ast}_{\alpha}(h) = 0$.

Now, for every $g:\lambda^{+} \rightarrow 2$, there are functions $\{ h_\gamma : \gamma \in \delta \} \subset \,^{\lambda^{+}}2$ and clubs $\{ C_\gamma: \gamma \in \delta \} \subset \mathcal{P}(\lambda^{+})$ such that, for each $\gamma \in \delta$,

$$ \{ \alpha \in S_\gamma \cap C_\gamma : F^{\gamma}_{\alpha}(h_\gamma\vert_\alpha) = g(\alpha) \} = \emptyset. $$

So, letting $D=\cap \{ C_\gamma : \gamma \in \delta \}$ and defining $h:\lambda^{+}\rightarrow 2$ by $h(\varphi(\gamma,\xi))=h_\gamma(\xi)$ for $\gamma \in \delta$, and $h(\varphi(\gamma, \xi)) = 0$ otherwise; we have $\alpha \in S_\gamma \cap C \cap D\implies $

$$ F^{\ast}_{\alpha}(h\vert_\alpha) = F^{\gamma}_{\alpha}([h]_{\gamma}\vert_\alpha)=F^{\gamma}_{\alpha}(h_{\gamma}\vert_\alpha)=1 - g(\alpha).$$
Therefore, $ \{ \alpha \in E\cap C \cap D: F^{\ast}_{\alpha}(h\vert_\alpha) = g(\alpha) \} = \emptyset $, and it immediately follows that $$E=\cup \{ S_\gamma: \gamma \in \delta \} \in I^{\lambda^{+}}_{WD}$$ $\square$.