Let's say I have the problem:
$$ \begin{cases} -\Delta u - b(x) \nabla u = f\\ u|_{\partial{\Omega}}=0 \end{cases} $$
With $f, \text{div}(b) \in L^2(\Omega)$ and $\Omega$ bounded. I would like to prove this function has a weak solution in $H_0 ^1 (\Omega)$. After applying integration by parts I should have that my PDE can be rewritten in weak form as:
$$ \int_{\Omega} \nabla u \nabla v = \int_{\Omega} (f + \text{div}(b))v \qquad \forall v \in H_0 ^1$$
Now, the LHS is a scalar product that induces an equivalent norm on $H_0 ^1$ thanks to Poincaré inequality, and the RHS can be viewed as a linear and continuous operator. Then Riesz's theorem should give me a unique solution. My question is: is this correct? And could you provide an example where instead of Riesz you have to use Lax-Milgram?
For your equation, you need extra assumption to use Lax-Milgram.
The bilinear you obtained is not quite right. The correct one should be as follows after you integrating by parts only for the second order term:
$$a(u, v):= (Lu,v) = \int_{\Omega}\big(\nabla u\cdot \nabla v -(b \cdot \nabla u) v\big) d x.\tag{1}$$
Lax-Milgram requires $$a(v,v) \geq \|v\|_{H^1(\Omega)}^2,$$ to get which, if we manipulate the advection term in (1) a little bit: $$ -\int_{\Omega}(b \cdot \nabla v) v d x=-\int_{\Omega} \nabla v\cdot (b v) d x = \int_{\Omega} v \operatorname{div} (b v) d x = \int_{\Omega} \operatorname{div} b\; v^{2} d x+\int_{\Omega}(b \cdot \nabla v) v d x, $$ yields $$ -\int_{\Omega}(b\cdot \nabla v) v d x=\frac{1}{2} \int_{\Omega} (\operatorname{div} b)v^{2} d x. $$ Hence in order to use Lax-Milgram, $\operatorname{div} b$ cannot be too negative.