Integration by parts (Green's identities)

Under the assumption that the condition $\Vert \boldsymbol{b}\Vert_{L^p} \leq C < \infty$ holds also for $p \to \infty$ you can show continuity for the bilinearform

\begin{align} B[u,v] := \int_U \nabla u \cdot \nabla v + v \boldsymbol{b} \cdot \nabla u \mathrm d x\end{align}

through \begin{align} \big \vert B[u,v] \big \vert =& \Bigg \vert \int_U \nabla u \cdot \nabla v + v \boldsymbol{b} \cdot \nabla u \mathrm d x \Bigg \vert = \Bigg \vert \int_U \nabla u \cdot \nabla v \mathrm d x + \int_U v \boldsymbol{b} \cdot \nabla u \mathrm d x \Bigg \vert \\ = & \Big \vert \big\langle \nabla u, \nabla v \big\rangle_{L^2(U)} + \big \langle v, \boldsymbol{b} \cdot \nabla u \big \rangle_{L^2(U)}\Big\vert \\ \overset{\text{Triangle-Ineq.}}{\leq} & \Big \vert \big\langle \nabla u, \nabla v \big\rangle_{L^2(U)} \Big \vert + \Big \vert \big \langle v, \boldsymbol{b} \cdot \nabla u \big \rangle_{L^2(U)}\Big\vert \\ \overset{\text{Cauchy-Schwarz}}{\leq} & \Vert \nabla u \Vert_{L^2(U)} \Vert \nabla v \Vert_{L^2(U)} + \Vert v \Vert_{L^2(U)} \Vert \boldsymbol{b} \cdot \nabla u \Vert_{L^2(U)} \\ \leq & \Vert \nabla u \Vert_{L^2(U)} \Vert \nabla v \Vert_{L^2(U)} + \Vert v \Vert_{L^2(U)} \big \Vert \Vert \boldsymbol{b} \Vert_{L^\infty(U)} \nabla u \big \Vert_{L^2(U)} \\ \leq & \Vert u \Vert_{W^{1,2}(U)} \Vert v \Vert_{W^{1,2}(U)} + \Vert v \Vert_{W^{1,2}(U)} \Vert \boldsymbol{b} \Vert_{L^\infty(U)} \Vert u \Vert_{W^{1,2}(U)} \\ \leq & \max \big \{1, \Vert \boldsymbol{b} \Vert_{L^\infty(U)} \big\} \Vert_{L^\infty(U)} \Vert u \Vert_{W^{1,2}(U)} \end{align}

Concerning your second assumption, $\nabla \cdot \boldsymbol{b} \geq 0$, I am not sure if you can show coercivity for this case. I am more familiar with the case $c -0.5 \nabla \cdot \boldsymbol{b} \geq 0 $ for the elliptic PDE $\Delta u + \boldsymbol{b} \cdot \nabla u + c u = f$, which would translate here ($c = 0$) to $\nabla \cdot \boldsymbol{b} \leq 0$.
In that case, you can show first $$\int_U u \boldsymbol{b} \cdot \nabla u \mathrm d x = \int_U \nabla \cdot (\boldsymbol{b}u^2) - u \boldsymbol{b}\cdot \nabla u - u^2 \nabla \cdot \boldsymbol{b} \mathrm d x $$ Thus, with $u \in W^{1, 2}_0(U)$ $$\int_U u \boldsymbol{b} \cdot \nabla u \mathrm d x = 0.5 \int_U - u^2 \nabla \cdot \boldsymbol{b} \mathrm d x \geq 0 $$ which would imply \begin{align} \big\vert B[u,u] \big\vert =& \Bigg \vert \int_U \nabla u \cdot \nabla u \mathrm d x + \int_U u \boldsymbol{b} \cdot \nabla u \mathrm d x \Bigg \vert \\ \geq& \Bigg \vert \int_U \nabla u \cdot \nabla u \mathrm d x \Bigg \vert = 0.5 \Bigg\vert \bigg[ \int_U \nabla u \cdot \nabla u \mathrm d x +\int_U \nabla u \cdot \nabla u \mathrm d x \bigg] \Bigg\vert \\ \overset{\text{Poincare-Friedrichs}}{\geq}& 0.5 \Big[ \Vert \nabla u \Vert^2_{L^2(U)} + \frac{1}{C} \Vert u \Vert^2_{L^2(U)} \Big] \\ \geq & 0.5 \min \{1, 1/C \} \Vert u \Vert^2_{W^{1, 2}(U)}\end{align}