There are various "high school" approaches to the answer. You will have to help me by drawing a picture.
Drop a perpendicular from the point $P$ to a point $M$ on the negative $x$-axis. Look at the angle $MOP$, and call it $\theta$. In $\triangle OPM$, the hypotenuse $OP$ has length $\sqrt{(\sqrt{3})^2+1}$, which is $2$. Thus $\sin\theta=1/2$.
You may recall "special angles." The angle $\theta$ between $0^\circ$ and $90^\circ$ such that $\sin\theta=1/2$ is the $30^\circ$ angle.
Now drop a perpendicular from $Q$ to the point $N$ on the positive $x$-axis. Let $\phi$ be the angle $QON$. What is the size of $\phi$? It is $180^\circ-(90^\circ+30^\circ)$, which is $60^\circ$. Thus $\phi$ is a lot bigger than the $30^\circ$ angle $\theta$, so the picture should not be at all symmetrical bout the $y$-axis!
Note that the cosine of the angle $\phi$ is $s/2$. But the cosine of the $60^\circ$ angle is $1/2$. It follows that $s=1$.
Without special angles: Do the constructions of $M$ and $N$ exactly as in the first solution, and let $\theta$, $\phi$ be as described there.
Note that $\theta+\phi=90^\circ$, so $\theta$ and $\phi$ are complementary angles.
Now compare $\triangle OPM$ and $\triangle QON$. We have $OP=QO$, and $\angle OPM=\angle QON$. So the triangles are congruent. Note that $PM=ON$. But $PM=1$, and therefore $s=ON=1$.
Using some analytic geometry: The slope of the line $OP$ is $-1/\sqrt{3}$. But $OQ$ is perpendicular to $OP$, so its slope is the negative reciprocal of $-1/\sqrt{3}$. Thus the slope of $OQ$ is $\sqrt{3}/1$.
But the slope of $OQ$ is $t/s$. It follows that $t=s\sqrt{3}$. By the Pythagorean Theorem, $s^2+t^2=4$. So $s^2+3s^2=4$. Since $s$ is positive, we conclude that $s=1$.