3 added 192 characters in body

In the presence of large cardinals, one can (or rather Shelah can...) force the answer to be "NO" in a very strong sense. The place to look is Section 7 of Chapter X of Proper and Improper Forcing.

In particular, Theorem 7.4 shows that assuming the consistency of 2 supercompact cardinals, one can force that for any regular $\kappa>\omega_1$, any stationary subset of $S^\kappa_{\aleph_0}$ contains a closed copy of $\omega_1$.

This implies the answer to your question is no by the following argument:

Step 1: If $\kappa>\aleph_1$ is regular and $A$ reflects at all uncountable limit ordinals below $\kappa$, then so does $A\cap S^\kappa_{\aleph_0}$ (where $S^\kappa_\tau$ is the set of ordinals less than $\kappa$ of cofinality $\tau$).

Proof: Let $A_0= A\cap S^\kappa_0$, and let $A_1= A\setminus A_0$. $A_1$ cannot reflect at ordinals of cofinality $\omega_1$, and so it must be the case that $A_0$ reflects at all ordinals of cofinality $\omega_1$. But then $A_0$ also reflects at any place where $S^\kappa_{\aleph_1}$ reflects as well, and so $A_0$ reflects at all ordinals of uncountable cofinality below $\kappa$.

Step 2:
Assume we are in a model like that obtained by Shelah. If $\kappa$ is a regular cardinal greater than $\aleph_1$ and $A$ is a stationary subset of $S^\kappa_{\aleph_0}$. We know $A$ contains a closed copy $C$ of $\omega_1$, and if we set $\delta=\sup(C)$ then $\delta$ is an ordinal of cofinality $\omega_1$ where $A$ reflects but $\kappa\setminus A$ does not. In particular, no stationary subset disjoint to $A$ can reflect at $\delta$, hence there is no way to get your "$B"$.

Edit:

A "no" answer to your question at $\omega_2$ is equiconsistent with the existence of a Mahlo cardinal.

As Joel mentioned in (an earlier version of) his answer, one can build $A$ and $B$ in $\omega_2$ from a $\square_{\omega_1}$-sequence. The failure of $\square_{\omega_1}$ implies that $\aleph_2$ is Mahlo in $L$ (Credited to Jensen on page 453 of Jech's "Set Theory"; I don't know a better reference.)

On the other hand, Theorem 7.1 in Chapter XI (page 576) of Proper and Improper forcing tells us that from a Mahlo cardinal, we can force ZFC+GCH + "every stationary subset of $S^{\omega_2}_{\omega}$ contains a closed copy of $\omega_1$, which we argued above gives a "No" answer.

Note that what Shelah is really showing is the consistency of the following statement:

"If $S$ is a stationary subset of $S^{\omega_2}_{\omega}$ that reflects at every member of $S^{\omega_2}_{\omega_1}$, then $S^{\omega_2}_{\omega}\setminus S$ is non-stationary.non-stationary,"

while the original question is equivalent to asking of $S^\kappa_\omega$ can be partitioned into two disjoint stationary sets, each of which reflects at every ordinal in $S^\kappa_{\omega_1}$.

A "no" answer to your question at $\omega_2$ is equiconsistent with the chapterexistence of a Mahlo cardinal.
As Joel mentioned in (an earlier version of) his answer, it seems one can build $A$ and $B$ in $\omega_2$ from a $\square_{\omega_1}$-sequence. The failure of $\square_{\omega_1}$ implies that obtaining $\aleph_2$ is Mahlo in $L$ (Credited to Jensen on page 453 of Jech's "Set Theory"; I don't know a better reference.)
On the other hand, Theorem 7.1 in Chapter XI (page 576) of Proper and Improper forcing tells us that from a Mahlo cardinal, we can force ZFC+GCH + "no" for every stationary subset of $\omega_2$ alone requires only S^{\omega_2}_{\omega}$contains a bit closed copy of measurability$\omega_1$, which we argued above gives a "No" answer. Note that what Shelah is really showing is the consistency of the following statement: "If$S$is a stationary subset of$S^{\omega_2}_{\omega}$that reflects at every member of$S^{\omega_2}_{\omega_1}$, then$S^{\omega_2}_{\omega}\setminus S$is non-stationary." 1 In the presence of large cardinals, one can (or rather Shelah can...) force the answer to be "NO" in a very strong sense. The place to look is Section 7 of Chapter X of Proper and Improper Forcing. In particular, Theorem 7.4 shows that assuming the consistency of 2 supercompact cardinals, one can force that for any regular$\kappa>\omega_1$, any stationary subset of$S^\kappa_{\aleph_0}$contains a closed copy of$\omega_1$. This implies the answer to your question is no by the following argument: Step 1: If$\kappa>\aleph_1$is regular and$A$reflects at all uncountable limit ordinals below$\kappa$, then so does$A\cap S^\kappa_{\aleph_0}$(where$S^\kappa_\tau$is the set of ordinals less than$\kappa$of cofinality$\tau$). Proof: Let$A_0= A\cap S^\kappa_0$, and let$A_1= A\setminus A_0$.$A_1$cannot reflect at ordinals of cofinality$\omega_1$, and so it must be the case that$A_0$reflects at all ordinals of cofinality$\omega_1$. But then$A_0$also reflects at any place where$S^\kappa_{\aleph_1}$reflects as well, and so$A_0$reflects at all ordinals of uncountable cofinality below$\kappa$. Step 2: Assume we are in a model like that obtained by Shelah. If$\kappa$is a regular cardinal greater than$\aleph_1$and$A$is a stationary subset of$S^\kappa_{\aleph_0}$. We know$A$contains a closed copy$C$of$\omega_1$, and if we set$\delta=\sup(C)$then$\delta$is an ordinal of cofinality$\omega_1$where$A$reflects but$\kappa\setminus A$does not. In particular, no stationary subset disjoint to$A$can reflect at$\delta$, hence there is no way to get your "$B"$. Glancing at the chapter, it seems that obtaining a "no" for$\omega_2\$ alone requires only a bit of measurability.