Can there be a point on a Riemann surface such that every rational function is ramified at this point?

If $S=\lbrace P\rbrace $, put an algebraic structure $X^{alg}$ on $X$ and take a uniformizing parameter $ t\in \mathcal O_{X^{alg},P}$ at $P$.
Since $\mathcal O_{X^{alg},P}\subset Rat(X^{alg})=\mathcal Mer(X)$, the meromorphic function $t$ seen as a map $X\to \mathbb P^1(\mathbb C)$ solves your problem.
You can generalize that to the case where $S$ is an arbitrary finite set, again by putting an algebraic structure on the Riemann surface $X$: look up Corollary 1.16 in Chapter VI of Miranda's book, which solves your problem.

Edit: Let me however give a self-contained proof using Riemann-Roch.

Fix an arbitrary point $x_0\in X\setminus S$ outside of $S=\lbrace x_1,...,x_n \rbrace$.
Consider the divisors (where $N$ will be determined later)
$$D_1=(-2)\cdot x_1+...+(-2)\cdot x_n+N\cdot x_0\quad \text {and} \quad D_2=(-1)\cdot x_1+...+(-1)\cdot x_n+N\cdot x_0$$ and their associated sheaves (=line bundles) $\mathcal O(D_1), \mathcal O(D_2)$.
They give rise to a short exact sequence of sheaves $$ 0\to \mathcal O(D_1)\to \mathcal O(D_2)\to \mathcal Q\to 0 $$ where the quotient sheaf $\mathcal Q$ is a finite sum of sky-scraper sheaves of rank $1$.
Taking cohomology we get $$ ...\to H^0(X, D_2)\to H^0(X,\mathcal Q) \to H^1(X,D_1)\to ...$$
Now $H^1(X,D_1)$ is dual to $H^0(X,K_X(2\cdot x_1+...+2\cdot x_n-N\cdot x_0))$ (by Serre duality) and is thus zero for $N\gt 2n+2g-2 $.
This choice of $N$ implies that the morphism $\gamma : H^0(X, D_2)\to H^0(X,\mathcal Q)$ is surjective.

And what has this got to do with your problem? It solves it!
Indeed, if you choose a coordinate $z_i$ near $x_i$, the stalk $\mathcal Q_{x_i}$ is identified with the complex line $\mathbb C\cdot z_i$ and by choosing a section $s\in H^0(X, \mathcal O(D_2))$ mapping to a $\gamma(s)\in H^0(X,\mathcal Q)$ non-zero at every $x_i$ you obtain the required meromorphic function $s$: its only pole is at $x_0$ and it has zeros of order exactly $1$ at the $x_i$'s.