We answer only the expectation question in the title, and not the more complicated distribution question asked in the body of the post. Questions like this one have been asked several times on MSE. There is also a large technical literature on related questions.
For $i=1$ to $N$, let random variable $X_i$ be $1$ if $i$ is chosen at least once, and let $X_i=0$ otherwise.
The probability that $X_i=1$ is $1$ minus the probability that the number is chosen no times. On any one trial, the probability of not choosing $i$ is $\frac{N-1}{N}$. Hence $\Pr(X_i=1)=1-\left(\frac{N-1}{N}\right)^k.$ The number $Y$ of $i$ chosen is given by $Y=\sum_{i=1}^N X_i,$ so by the linearity of expectation, $E(Y)=\sum_{i=1}^N X_i=N\left( 1-\left(\frac{N-1}{N}\right)^k \right).$ For the expected proportion, divide by $N$.