Your question is about the vectors in $V_1\otimes V_2$ that provide $1$-dimensional $B_\Delta$-subrepresentations of weight $\mu$, where $B_\Delta$ is the diagonal in $B\times B \leq G\times G$. Let's call this subspace $J$.

Let $S \hookrightarrow T$ be a regular dominant coweight, and let $B_z := (S(z)\times 1)\ B_\Delta\ (S(z^{-1})\times 1)$. Then $B_1 = B_\Delta$, and $\lim_{z\to\infty} B_z = (B' \times 1) T_\Delta$, where $B'\leq B$ is the unipotent radical. We can also consider $J_z := (S(z)\times 1)\cdot J$, and its limit $J_\infty$ (computed inside the Grassmannian of $(\dim J)$-dimensional subspaces of $V_1\otimes V_2$).

Now I claim that $J_z$ is $B_z$-invariant for each $z$, and hence, $J_\infty$ is invariant under $(B'\times 1) T_\Delta$. Your assumption was that $J$ was not $0$; hence there are vectors of $(B'\times 1) T_\Delta$-weight $\mu$. Call that potentially larger space $J'$.

That limit group is now normalized by $T\times T$, which will act on $J'$ and break it into weight spaces, i.e. weight spaces of $V_1$ tensor weight spaces of $V_2$. The $(B'\times 1)$-invariance means that the only $V_1$ weight space met this way is the high weight space.

Summing up, the assumption that $\mu$ is a high weight in $V_1 \otimes V_2$ implies that $\mu$ can be written as the high weight of $V_1$ plus a weight of $V_2$.

One way to rephrase your question was that you were looking for asymmetry. I get it here with this $S(z)\times 1$, whereas Friedrich Knop's crystal-based answer implicitly derives it from the quantum group.

Your question is about the vectors in $V_1\otimes V_2$ that provide $1$-dimensional $B_\Delta$-subrepresentations of weight $\mu$, where $B_\Delta$ is the diagonal in $B\times B \leq G\times G$. Let's call this subspace $J$.

Let $S \hookrightarrow T$ be a regular dominant coweight, and let $B_z := (S(z)\times 1)\ B_\Delta\ (S(z^{-1})\times 1)$. Then $B_1 = B_\Delta$, and $\lim_{z\to\infty} B_z = (B' \times 1) T_\Delta$, where $B'\leq B$ is the unipotent radical; call this limit subgroup $B_\infty$. We can also consider $J_z := (S(z)\times 1)\cdot J$, and its limit $J_\infty$ (computed inside the Grassmannian of $(\dim J)$-dimensional subspaces of $V_1\otimes V_2$).

Now I claim that $J_z$ is $B_z$-invariant for each $z$, and hence, $J_\infty$ is invariant under $(B'\times 1) T_\Delta$. Your assumption was that $J$ was not $0$; hence there are vectors of $(B'\times 1) T_\Delta$-weight $\mu$. Call that potentially larger space $J'$.

Unlike $B_\Delta$, the limit group $B_\infty$ is normalized by $T\times T$, which will act on $J'$ and break it into weight spaces, i.e. weight vectors of $V_1$ tensor weight vectors of $V_2$. The $(B'\times 1)$-invariance means that the only $V_1$ weight space met this way is the high weight space.

Summing up, the assumption that $\mu$ is a high weight in $V_1 \otimes V_2$ implies that $\mu$ can be written as the high weight of $V_1$ plus a weight of $V_2$.

One way to rephrase your question was that you were looking for asymmetry. I get it here with this $S(z)\times 1$, whereas Friedrich Knop's crystal-based answer implicitly derives it from the quantum group (for which $V_1\otimes V_2$ is not obviously isomorphic to $V_2\otimes V_1$, especially as $q\to 0$).