As you know, when $\mathrm P(B)\ne0$, $\mathrm P(A\mid B)=\mathrm P(A\cap B)/\mathrm P(B)$. The trouble in your context is that $\mathrm P(B)=0$, and even, as you say, that $\mathrm P(A)=0$. A way to define nevertheless some quantity $\mathrm P^*(A\mid B)$ akin to $\mathrm P(A\mid B)$ is to replace $A$ and $B$ by some sets $A_t$ and $B_t$ whose probabilities are positive for every positive $t$ and such that, in a sense, $A_t\to A$ and $B_t\to B$ when $t\to0$. Then one could compute $\mathrm P(A_t\mid B_t)$ for every positive $t$ in the usual way and see if this quantity has a limit when $t\to0$. If so, the limit could be chosen as $\mathrm P^*(A\mid B)$.
In the case $A=\{a\}$ and $B=\{a,b\}$ with $a$ and $b$ in $(0,1)$, one can consider $A_t=A+[-t,t]$ and $B_t=B+[-t,t]$, that is, $A_t=[a-t,a+t]$ and $B_t=[a-t,a+t]\cup[b-t,b+t]$.
Assume that $t$ is small enough. Then, $[a-t,a+t]\subset[0,1]$ hence $\mathrm P(A_t)=2t$, and $[a-t,a+t]\cup[b-t,b+t]\subset[0,1]$ with $[a-t,a+t]\cap[b-t,b+t]=\varnothing$, hence $\mathrm P(B_t)=4t$. Thus $\mathrm P(A_t\mid B_t)=\frac12$ for every $t$ small enough, which suggests indeed that $\mathrm P^*(A\mid B)=\frac12$ is a reasonable choice.
Note that this procedure is relatively robust since $[-t,t]$ could be replaced by any neighbourhood of $0$ shrinking to $\{0\}$ when $t\to0$, for example $[-2t,5t+t^4]$, without changing the final result.
More generally, assume that $A=\{a_1\}$ and $B=\{a_1,a_2,\ldots,a_n\}\subset(0,1)$, that $\mathrm P$ has density $f$ and that $f$ is continuous at $a_k$ for every $1\leqslant k\leqslant n$. One sees that the reasoning above suggests to choose $\mathrm P^*(A\mid B)=\frac{f(a_1)}{f(a_1)+\cdots+f(a_n)}$. In effect, this is equivalent to replacing the nonexistent conditional probability $\mathrm P(\ \mid B)$ by the discrete probability measure $\mathrm P^*(\ \mid B)=\mu_B$ defined by $\mu_B(\{a_k\})=\frac{f(a_k)}{f(a_1)+\cdots+f(a_n)}$ for every $k$ and $\mu_B(C)=0$ if $B\cap C=\varnothing$.