Consider the weak form using the SUPG method: $$(u_t, v+\delta b \cdot \nabla v) + a(u,v) = (f, v+\delta b \cdot \nabla v ),$$ where $a(u,v) = (b \cdot \nabla u, \delta b \cdot \nabla v )$. How to prove the well-posedness of this weak form.
If we have test function $v$ instead of $\delta b \cdot \nabla v$, one can use a coercivity-like argument. As follows: By multiplying test function $v = u$ in above equation to obtain that \begin{equation}\label{stfe20} (u_{t},u+\delta b \cdot \nabla u)+a(u,u+\delta b \cdot \nabla u)=(f,u+\delta b \cdot \nabla u), \end{equation} which is reduced to \begin{align}\label{stfe21} \dfrac{1}{2}\dfrac{d}{dt}\|u\|^{2}_{L^{2}(\Omega)}+\delta\|b \cdot \nabla v\|^{2}_{L^{2}(\Omega)} +\||b \cdot \mathbf{n}|^{1/2}v\|^{2}_{L^{2}(\partial \Omega)}\\[4pt] \leq \left|(f,u+\delta b \cdot \nabla u)\right|+\left|\delta(u_{t}, b \cdot \nabla u)\right|, \end{align} Next, we estimate right-hand side terms as follows: \begin{align}\label{stfe202} \left|(f,u)\right| \leq 2\|f\|^{2}_{L^{2}(\Omega)} + \dfrac{1}{8} \|u\|^{2}_{L^{2}(\Omega)} \end{align} and \begin{align}\label{stfe203} \left|\delta(f, b \cdot \nabla u)\right| \leq 2\delta\|f\|^{2}_{L^{2}(\Omega)} + \dfrac{\delta}{8} \|b \cdot \nabla u\|^{2}_{L^{2}(\Omega)}. \end{align} In a similar way, the second term is bounded by \begin{align}\label{stfe204} \left|\delta(u_{t}, b \cdot \nabla u)\right| \leq 2\delta\|u_{t}\|^{2}_{L^{2}(\Omega)} + \dfrac{\delta}{8} \|b \cdot \nabla u\|^{2}_{L^{2}(\Omega)}. \end{align} Now, how we do manage this last term $\delta\|u_{t}\|^{2}_{L^{2}(\Omega)}$, while using all RHS terms in original bilinear form? All other terms are manageable.