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 $$s(x)-x=\alpha\quad \quad (1)$$ is quadratic with respect to $x$, thus it has even number of roots unless it has a double root. We prove that there exists unique quadratic residue $\alpha$ for which (1) has a double 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 a fixed point, we have $c\ne 0$ (else $\infty$ would be a fixed point) and the discriminant $$(d-a)^2+4bc=(d+a)^2-4=(d+a-2)(d+a+2)$$ of the equation $ax+b-x(cx+d)=0$ is not a square. Therefore one of the numbers $d+a-2,d+a+2$ is a quadratic residue and the other is not. Equation (1) 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.