Matching problem in continuous time Let's split each period into $n$ intervals. There's a continuum $u$ of unemployed and $v$ of vacancies. During each interval, there is a total of $X/n$ job offers. That means that each unemployed gets a job offer with probability $X/(nu)$ (assuming that $n$ is eventually so small that the chance of multiple job offers to the same person will go to zero). 

Single individual Number of job offers - for a single individual - in the interval is distributed $Binomial(k, n, \frac{X}{un})$. Letting $n\to\infty$, the distribution in the continuous-time analog converges to $Poisson(k, \frac{X}{u})$. 

I'm interested in the probability of $x$ individuals getting at least one job interval during a whole interval. That is $(1 - Binomial(0, n, \frac{X}{nu}))^x$:

$$ (1 - (1 - \frac{X}{un})^n)^x $$

If I'm not mistaken, the continuous-time analog is 

$$ (1 - e^\frac{-X}{u})^x$$

Is that correct? The combination of exponential and power function is making me quite uncomfortable.
An alternative approximating approach you could use as as check might be to say there are $X$ job offers in total and $u$ unemployed. 

So the probability that an individual does not get a particular job offer is $\left(1-\dfrac{1}{u}\right)$ and so the probability the individual does not get any job offer is $\left(1-\dfrac{1}{u}\right)^X$ which is $\left(\left(1-\dfrac{1}{u}\right)^u\right)^{x/u} \approx e^{-X/u}$ much as one might expect from a Poisson distribution. You have an error in the sign of $e^{-X/u}$.

That makes the probability the individual does get a job offer about $1-e^{-X/u}$. It makes the expected number of individuals receiving job offers about $u(1-e^{-X/u})$.  

The probability that $y$ individuals get job offers (rather than $x$ as you have already used $X$) is then roughly the binomial $\displaystyle {y \choose u}\left(1-e^{-X/u}\right)^y e^{-(u-y)X/u}$ and or if you wanted a Poisson approximation to this then about $\dfrac{e^{-u\left(1-e^{-X/u}\right)}u^y\left(1-e^{-X/u}\right)^y}{y!}$ and you could apply further approximations to this but it would not get much tidier