It may help to rewrite your integrand as
$ y_s(x) = \frac{3}{4s^3}(s^2 - x^2) = \frac 1s \frac 34 \left( 1 - \left( \frac xs \right)^2 \right). $
Now, let's first consider the case $s = 1$, which simplifies the integrand to
$ y_1(x) = \frac 34 (1 - x^2), $
which is to be integrated from $-1$ to $1$. From high school calculus, I'm sure you'll remember that
$ \int_0^1 x^2 \,dx = \frac 13 $
which implies that
$ \int_0^1 (1 - x^2) \,dx = 1 - \frac 13 = \frac 23 $
and thus, since $(-x)^2 = x^2$,
$ \int_{-1}^1 (1 - x^2) \,dx = 2 \cdot \frac 23 = \frac 43. $
Multiply that by $\frac 34$, and you get $1$.
Now, what about $s \ne 1$? The rewritten integrand has a factor of $\frac 1s$ in front of it, so the height of the parabola is divided by a factor of $s$. On the other hand, the argument $x$ is also divided by $x$, which effectively multiplies the width of the parabola (and, not coincidentally, the range over which it is integrated) by $s$. Those two effects cancel out, leaving the area unchanged.
In fact, we can generalize this result: for any function $f$ integrable on $[a,b]$ and any non-zero constant $s$,
$\begin{aligned} \int_a^b f(x) dx &= \frac 1s \int_{sa}^{sb} f\left(\frac xs\right) \,dx \\ &= \int_{sa}^{sb} \frac 1s f\left(\frac xs\right) \,dx . \end{aligned}$