In the Monty Hall problem, why isn't Pr(I switch and win| Monty opens door 2) = 1/2?

You were close.

The Law of Total Probability will actually state: $$\small\begin{align}\mathsf P(S\mid M_2)&=\mathsf P(S\mid M_2,D_1)\,\mathsf P(D_1\mid M_2)+\mathsf P(S\mid M_2,D_3)\,\mathsf P(D_3\mid M_2)\\[1ex]&=0\cdot\mathsf P(D_1\mid M_2)+1\cdot\mathsf P(D_3\mid M_2)\\[1ex]&=\mathsf P(D_3\mid M_2)\end{align}$$

(Which is sensible: the conditional probability that switching wins when door two is revealed, equals the conditional probability that the prise is behind the third door when door two has been revealed).

Now when Monty picks door 2, the prise does not have an equal probability for being behind door 1 and 3. There are two possible outcomes, but they are not equally likely. Monty is more likely to have picked door 2 if the prise is behind door 3 than if it is behind door 1.

However, we can evaluate this by applying Bayes' Theorem ... $$\small\begin{align}\mathsf P(D_3\mid M_2)&=\dfrac{\mathsf P(D_3)\,\mathsf P(M_2\mid D_3)}{\mathsf P(M_2)}\\[1ex]&=\dfrac{\mathsf P(D_3)\,\mathsf P(M_2\mid D_3)}{\mathsf P(D_1)\,\mathsf P(M_2\mid D_1)+\mathsf P(D_3)\,\mathsf P(M_2\mid D_3)}\\[1ex]&=\dfrac{1}{\tfrac 12+1}\\[1ex]&=\dfrac{2}{3}\end{align}$$

When the prise is behind door 1, Monty may open either door 2 or 3, if we assume he does so without bias:$$\small\mathsf P(M_2\mid D_1)=\dfrac 12$$

When the prise is behind door 3, Monty may only open door 2:$$\small\mathsf P(M_2\mid D_3)=1$$