During a given year, each family either (1) has a unique child or (2) has no child. Those who previously had a boy decide to have no supplementary child, hence these are all in case (2), but maybe some others are in case (2) as well, for other reasons of their own, this does not matter.
What matters is that each family in case (1) has as much chances to have a boy than a girl. By the law of large numbers, if the number $M$ of families who do procreate during this given year is large and if each procreates independently on the others, $\frac12M+r_M$ boys are born and $\frac12M-r_M$ girls are born, where $r_M$ is random and $|r_M|\ll M$. The proportion of boys amongst the children born this year is $\frac{M/2+r_M}M=\frac12+\varepsilon_M$ with $\varepsilon_M=\frac{r_M}M$ hence $|\varepsilon_M|\ll1$. In other words, roughly one half of all the children born this year are boys.
Thus the hypothesis that the global population is large is important, but the details of the strategy (in the present case, Stop after one boy) are simply not relevant since every adapted strategy (in the sense that the decision on a given year only depends on what happened on the previous years) would yield the same result.
Edit (This is to answer a comment by the OP.)
The preceding paragraphs describe the almost sure behaviour in the limit of large initial populations. Turning to the behaviour in the mean for finite initial populations, note that the distribution of $r_M$ is symmetric, since $r_M$ is the sum of a random number $M$ of i.i.d. centered $\pm1/2$ Bernoulli random variables. Hence $\mathrm E\left(\frac{b_k}{g_k+b_k}\right)=\frac12$ exactly, where $b_k$ and $g_k$ denote the numbers of boys and girls born in generation $k$.
This does not imply that the total numbers $B_k=b_1+\cdots+b_k$ and $G_k=g_1+\cdots+g_k$ of boys and girls born until generation $k$ fulfill the same property.
Consider for example the second generation. Then the distribution of $b_1$ is binomial $(N,\frac12)$, $g_1=N-b_1$, the conditional distribution of $b_2$ conditionally on $b_1$ or $g_1$ is binomial $(g_1,\frac12)$, and $g_2=g_1-b_2$. In particular, $\mathrm E(b_1)=\frac12N$ and $\mathrm E(b_2\mid b_1)=\frac12(N-b_1)$.
Consider the successive ratios $R_k=\frac{B_k}{B_k+G_k}$. Then $R_1=\frac{b_1}N$ hence $\mathrm E(R_1)=\frac12$. On the other hand, $R_2=\frac{b_1+b_2}{N+g_1}=\frac{b_1+b_2}{2N-b_1}$ hence $\mathrm E(R_2\mid b_1)=\frac{b_1+(g_1/2)}{2N-b_1}=\frac12\frac{N+b_1}{2N-b_1}$. By convexity, $ \mathrm E(R_2)\gt\frac12\frac{N+\mathrm E(b_1)}{2N-\mathrm E(b_1)}=\frac12\frac{N+(N/2)}{2N-(N/2)}=\frac12, $ hence $\mathrm E(R_2)\ne\frac12$.
Second edit
Counting the children family by family instead of generation by generation, one sees readily that $R_k\to R_\infty$ almost surely when $k\to\infty$, where $R_\infty=\frac{N}{N+\sigma_N}$ and $\sigma_N$ the sum of $N$ i.i.d. geometric random variables $\tau_i$ of parameter $\frac12$, such that $\mathrm P(\tau_i=n)=2^{-n}$ for every $n\geqslant0$. Further computations then show that $ \mathrm E(R_\infty)=N\int_0^1\frac{u^{N-1}}{(2-u)^N}\mathrm du=\frac12+\frac34\frac1N+o\left(\frac1N\right). $ In particular, $\mathrm E(R_k)\ne\frac12$ for every $k$ large enough (and probably for every $k\geqslant2$).