This is to complete the very nice argument by Peter Mueller.

Denote $\Omega:=P^1\mathbb{F}_q=\mathbb{F}_q\cup \infty$, so that $|\Omega|=q+1$ and the group $G=PSL(2,q)$ acts on $\Omega$ by projective transformations. Assume that $S\subset G=PSL(2,q)$, $|S|=q+1$ is chosen so that 
$$\sum_{s\in S} \mathbb{1}_{s(i)=j}=1,\quad\forall i,j\in \Omega.$$
Then the same holds for the set $g_0S$ for any $g_0\in G$, thus we may suppose that $id\in S$, therefore other elements of $S$ do not have fixed points. Denote by $G^\star$ the set of fixed-point-free elements from 𝐺 together with  1, we have $S\subset G^\star$. Consider the function $M(i,j)$ on $\Omega\times \Omega$ defined as 
$$
M(i,j)=\begin{cases}1,\, \text{if}\,\, i=\infty,\, j\ne \infty\\
1,\, \text{if}\,\, i,j\in \mathbb{F}_q,\chi(i-j)=1\\
0,\, \text{otherwise}.\end{cases}
$$
Here $\chi$ is a quadratic character of $\mathbb{F}_q$ (Legendre symbol if $q$ is prime).
We get
$$
\sum_{i,j\in \Omega,s\in S} M(i,j)\mathbb{1}_{s(i)=j}=\sum_{i,j\in \Omega} M(i,j)=q+q(q-1)/2
$$
is odd. Thus to get a contradiction it suffices to prove that 
$$
\sum_{i,j\in \Omega} M(i,j)\mathbb{1}_{s(i)=j}
$$
is even for any fixed element $s\in G^\star$. For $s=id$ all summands are just zeroes. If $s$ does not have fixed point, than 
$$
\sum_{i,j\in \Omega} M(i,j)\mathbb{1}_{s(i)=j}=1+\sum_{i\in \mathbb{F}_q} M(i,s(i))=1+\left|i\in \mathbb{F}_q:\chi(s(i)-i)=1\right|.
$$
Note that for fixed $\alpha\in \mathbb{F}_q$, the equation $s(i)-i=\alpha$ is quadratic with respect to $i$, thus have even number of roots unless it has a multiple root. We prove that there exists quadratic residue $\alpha$ for which this equation has a multiple root, the result would follow.

Possibly this have some clever explanation, but at the moment I do not see it, so just calculate. Denote $s(x)=(ax+b)/(cx+d)$, $ad-bc=1$. Since $s$ does not have fixed point, we have $c\ne 0$ (else $\infty$ would be a fixed point) and the discriminant $$(d-a)^2+4bc=(d+a)^2-4$$
of the equation $ax+b=x(cx+d)$ is not a square. Therefore one of numbers $d+a-2,d+a+2$ is a quadratic residue and the other is not. We look for quadratic residues $\alpha$ such that the equation $s(x)-x=\alpha$ has a double root. This equation simplifies as
$$
(ax+b)-(cx+d)(x+\alpha)=-cx^2+x((a-d)-c\alpha)+b-d\alpha,
$$
thus it has a double root when the discriminant is zero:
$$
(a-d-c\alpha)^2+4c(b-d\alpha)=(\alpha c-(a+d))^2-4(ad-bc)=(\alpha c-(a+d+2))(\alpha c-(a+d-2))=0
$$
This happens for $\alpha=c^{-1}(a+d\pm 2)$, and exactly for one choice of the sign $\alpha$ appears to be a quadratic residue.