This is true, with the caveat that the semi-norm need not be one from whatever collection of seminorms was originally used to define the topology on $S$. Also, this is true for any locally convex space; it need not be a Fréchet space (which requires in addition a compatible complete translation-invariant metric).
Claim: In a locally convex space $X$ whose topology is defined by seminorms $\{p_\alpha\}$, a linear functional $f:X\to\mathbb{R}$ is continuous if and only if it is majorized by a finite combination of seminorms: $$|f(x)|\le M\max (p_1(x),\dots, p_n(x))$$
Proof: The set $U=\{x : |f(x)|<1\}$ is an open set containing $0$. By the definition of LCS topology, $U$ it contains a neighborhood of $0$ the form $\{x \colon \max_{k=1, \dots, n} p_k(x)<\epsilon\}$. The conclusion with $M=1/\epsilon$. $\quad \Box$
It remains to left $p(x) = \max (p_1(x),\dots, p_n(x))$ and you get the desired seminorm.
Example
This is an example to show that one can't always pick just one of "original" seminorms. Let $S$ be the space $C(\mathbb{R})$ of all continuous functions on $\mathbb{R}$ equipped with seminorms
$$
p_n(f) = \sup_{[n, n+1]}|f|,\qquad n\in\mathbb{Z}
$$
These seminorms induce the topology of uniform convergence on compact sets. The linear functional $T(f) = \int_{0}^3 f(x)\,dx$ is continuous on $S$, but there is no single seminorm $p_n$ that dominates $|T(f)|$; one has to use $\max(p_0, p_1, p_2)$.