Product of Bruhat Cells

Here is a (probably equivalent) way to see this. I'll give an argument for $(BsB)(BwB)$, where $s$ is a simple reflection. You should be able to expand this argument to $(BvB)(BwB)$ as requested in the question. The expansion will use the fact that $\ell(vw) = \ell(v) + \ell(w)$ implies $\ell(s_t \cdots s_1 w) = t + \ell(w)$, where $s_n \cdots s_1 = v$ is a reduced decomposition for $v$ and $n \geq t$.

Throughout, let $T\subset B$ be a fixed maximal torus and $U \subset B$ the unipotent subgroup of $B$. Recall that $B = TU$. For any positive root $\alpha$, let $U_\alpha$ be the corresponding root subgroup of $U$. Recall that for any ordering of positive roots, we can write $U$ as a produce of $U_\alpha$ in the chosen order. Fix an odering of positive roots so that $U = U_{w{-1}} U_{w_0w^{-1}}$, where $w_0$ is the longest element of the Weyl group.

Then, the elements of $BwB$ are uniquely written as $U_{w^{-1}}wB$, where \begin{equation} U_{w^{-1}} = \prod_{\alpha \in R(w^{-1})} U_\alpha. \end{equation} Here, $R(w^{-1}) =\{\alpha \in R^+\;|\; w^{-1} \alpha <0\;\}$. This decomposition follows from the Bruhat decomposition because if $w^{-1} \alpha >0$, then $w^{-1} u_{\alpha} w = u_{w^{-1} \alpha}$, and so $ u_{\alpha} w = wu_{w^{-1} \alpha}$ and $u_{w^{-1} \alpha} \in B$. So, this element is absorbed to the right.

Then, every element $ b s b_1b_2 w \tilde{b} \in (BsB)(BwB)$ can be written in the form $BsuwB$, where $u \in U_{w^{-1}}$ as above. Explicitly, $b_1b_2 = t(uu') \in B$, where $t \in T$ and $(uu')\in U_{w{-1}} U_{w_0w^{-1}}$ according to the decomposition given above. Then, $ t(uu')w \tilde{b} = u b'$ for some $u \in U_{w^{-1}}$ and $b' \in B$. For each $(b_1b_2)$, we get a unique $u b'$.

If we can show that $sR(w^{-1}) \subset R^+$, then for $u = u_{\alpha_1} \cdots u_{\alpha_k}$ (i.e, $\alpha_i \in R(w^{-1})$) we have \begin{equation} su = s (u_{\alpha_1} \cdots u_{\alpha_k})=(u_{s\alpha_1} \cdots u_{s\alpha_k})s, \end{equation} and $(u_{s\alpha_1} \cdots u_{s\alpha_k}) \in B$. Then, every $b s b_1b_2 w \tilde{b}= b(u_{s\alpha_1} \cdots u_{s\alpha_k})sw b'$. This tells us that every $b_3 sw b_4 = b(u_{s\alpha_1} \cdots u_{s\alpha_k})sw b'$ (sorry for the excessive indexing), where $(b_1b_2)$ can be arbitrary because we need only adjust $b$ and $b'$ to make this equation work. However, once $(b_1b_2)$ is fixed, $b$ and $b'$ are determined by this value.

To show $s R(w^{-1}) \subset R^+$, we use the assumption that \begin{equation} 1+ \ell(w^{-1}) = 1 + \ell(w) = \ell(sw) = \ell(w^{-1}s) . \end{equation} So, $|R(w^{-1}s)| = |R(w^{-1})| + 1$ since the cardinality of $R(w)$ is just the length of $w$. Using a basic formula for $R(w^{-1}s)$ (see, e.g., Springer's Linear Algebraic Groups, section 8.3.1) this implies $s R(w^{-1}) \subset R^+$ as desired.