Here is I what I have found based on a problem on Polya's urn scheme
but I elaborate this formally below
Proof for the specific case
We start with $7$ white balls and $13$ black balls. On each round we add additional $2$ balls according to the rules described in the question.
Think that for each round you either draw a ball that was there in the previous round or you draw a ball that was added in the previous round.
Assume inductively that on round $n$ the probability of drawing a black ball is $p$ ($:= \frac{b}{b+w}$). Let's see what happens on round $n+1$. Let $k$ denote the number of balls in the $n+1$ round. So with probability $\frac{k-2}{k}$ we draw an old ball (that was there in the $n$th round) and with probability $\frac{2}{k}$ we draw a new ball (one of the two that were added on the $n$th round). By the inductive hypothesis, the probability of drawing a black ball from the old ones is $p$. The probability of drawing a black ball from the new ones is simply $p$ (because you either have drawn a black ball with probability $p$ on the $n$th round and hence necessarily a black one on the $n+1$th round, or you have drawn a white one on the $n$th round and hence a white one on the $n+1$th round).
All in all, the probability of drawing a black ball on the $n+1$th round is:
$$p \cdot \frac{k-2}{k} + p \cdot \frac{2}{k} = p \cdot ( \frac{k-2}{k} + \frac{2}{k}) = p$$
Proof for the general case
You can use this idea to give a general inductive proof for the number $k$ of balls that you add each time to the urn and arbitrary initial configuration $b,w$ of black and white balls, respectively.
Let $b,w$ be our starting configuration.
On the first round the probability of drawing a black ball is $\frac{b}{b+w}$.
Now, assume inductively that on the $n$th round the probability of drawing a black ball is $\frac{b}{b+w}$. The probability of drawing a black ball on the $(n+1)$th round is:
$$P(B_{n+1}) = P(B_n) P(B_{n+1} | B_n) + P(W_n) P(B_{n+1} | W_n) ,$$
where $B_i$, $W_i$ are the events of drawing a black ball and the white ball on the $i$th round, respectively.
We have:
$$P(B_{n+1}) = \frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{w}{b+w}\frac{(b+w+(n-1)k)\frac{b}{b+w}}{b+w+nk}=$$
$$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{w}{b+w}\frac{b}{w}\frac{(b+w+(n-1)k)\frac{b}{b+w}\frac{w}{b}}{b+w+nk}=$$
$$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k}{b+w+nk} + \frac{b}{b+w}\frac{(b+w+(n-1)k)\frac{w}{b+w}}{b+w+nk}=$$
$$\frac{b}{b+w} \frac{(b+w + (n-1)k)\frac{b}{b+w} + k + (b+w+(n-1)k)\frac{w}{b+w}}{b+w+nk}$$
$$\frac{b}{b+w} \frac{(b+w + (n-1)k)(\frac{b}{b+w} + \frac{w}{b+w}) + k}{b+w+nk}$$
$$\frac{b}{b+w} \frac{(b+w + (n-1)k) + k}{b+w+nk}$$
$$\frac{b}{b+w} \frac{b+w + nk}{b+w+nk} = \frac{b}{b+w}$$
This ends the proof.