Observe that every commutative C*-algebra $A \not\simeq \mathbb{C}$ has a non-trivial zero-divisor. To see this view $A$ as continuous functions over its spectrum which contains at least two points (and is Hausdorff) and use Urysohn lemma to construct two continuous functions supported at separating open sets of these two points. It follows that every prime ideal is the kernel of a homomorphism to $\mathbb{C}$.

Now, given a multiplicative seminorm $s$ on $A$, observe that its kernel $p$ is a prime ideal and it defines a multiplicative norm on $A/p\simeq \mathbb{C}$. This norm must coincide with the usual absolute-value $|\cdot|$. Denoting the map $A\to\mathbb{C}$ by $a\mapsto a(x)$ we conclude that for every $a\in A$, $s(a)=|a(x)|$.

Note that the commutativity assumption was inessential in all of the above.

Observe that every commutative C*-algebra $A \not\simeq \mathbb{C}$ has a non-trivial zero-divisor. To see this view $A$ as continuous functions over its spectrum which contains at least two points (and is Hausdorff) and use Urysohn lemma to construct two continuous functions supported at separating open sets of these two points. It follows that every closed$^1$ prime ideal is the kernel of a homomorphism to $\mathbb{C}$.

Now, given a bounded multiplicative seminorm $s$ on $A$, observe that its kernel $p$ is a closed prime ideal and it defines a multiplicative norm on $A/p\simeq \mathbb{C}$. This norm must coincide with the usual absolute-value $|\cdot|$. Denoting the map $A\to\mathbb{C}$ by $a\mapsto a(x)$ we conclude that for every $a\in A$, $s(a)=|a(x)|$.

Note that the commutativity assumption was inessential in all of the above.

1: Thanks to Robert Furber for correcting my omission here.