Suppose $P(Y=y)=0$. Is $P(X=x|Y=y)$ a well defined object? Consider two discrete random variables $X$ and $Y$ defined on the same probability space $(\Omega, \mathcal{F}, P)$. Suppose $P(Y=y)=0$. Is $P(X=x|Y=y)$ a well defined object? 
I would say that the answer is NO given the formula
$$
P(X=x|Y=y)=\frac{P(X=x, Y=y)}{P(Y=y)}
$$
However, I remember some definition of conditional probability using measure theory and I'm wondering whether that would give a different answer. 
 A: In general conditional probabilities are only define almost everywhere. That is, if you have a joint probability measure $\mathsf P$ and $\mu$ is its first marginal (e.g. distribution of $X$ in your notation), then conditional distribution is any stochastic kernel $\kappa$ satisfying
$$
  \mathsf P(A\times B) = \int_A \kappa(B|x)\mu(\mathrm dx) \tag{1}
$$
for all measurable sets $A,B$. Due to the integral in condition on $\kappa$, it is not uniquely defined when exists. Namely, if $\kappa$ solves $(1)$ and $\mu(x = x_0) = 0$, then by changing $\kappa$ on $x_0$ we still get the solution of $(1)$. In contrast, if $\mu(x = x_0) \neq 0$ we get from $(1)$:
$$
  \mathsf P(A\times B) = \kappa(B|x_0)\mu(x = x_0) \quad\implies\quad \kappa(B|x_0) = \frac{\mathsf P(A\times B)}{\mu(x = x_0)}.
$$
Thus, if $\mu(x = x_0) > 0$  then the conditional probability is uniquely defined, but otherwise it can take any value at this particular point, whereas its average behavior must comply with $(1)$. That is, for every single point of zero measure, we can assign any value we want, but we can't do it to all of them at the same time, since a union of this points may have a non-zero measure.
So, short answer to your question: this quantity is not uniquely defined, but you can in fact talk of probability conditioned on zero event. With caution. 
A: That IS the definition using measure theory; $P$ is a measure! The answer is NO, for the reason you have given.
