$(0)$ is not always in $\operatorname {Spec} A$ is where your reasoning went wrong. One can show rather straightforwardly that $\operatorname {Spec} A$ is irreducible iff the the nilradical is prime (Exercise 19 in Ch. 1 of Atiyah and Macdonald's commutative algebra book) Note there is a natural homeomorphism induced by the projection map between $\operatorname {Spec} A/I$ and the closed subset of $\operatorname {Spec} A$, $V(I)= \{p\in \operatorname {Spec} A\ | \ I\subseteq p\}$, so this deals with the irreducible components question as well.
Just friendly advice, I would recommend you do most of the exercises in a book like Atiyah and Macdonald before seriously attempting to learn about schemes.
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=\operatorname{Spec}A$ and $Y=\operatorname{Spec}B$; one can prove that the function
$$
\Phi:(f,f^{\sharp})\in\operatorname{Hom}_{\bf LocRingSp}((X,\mathcal{O}_X),(Y,\mathcal{O}_Y))\to f^{\sharp}(Y)\in\operatorname{Hom}_{\bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{\sharp})$ is uniquely determined by $f^{\sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{\sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{\sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{\sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=\operatorname{Spec}A$; one can prove that the function
$$
\Phi:(f,f^{\sharp})\in\operatorname{Hom}_{\bf Sch}(Y,X)\to f^{\sharp}(X)\in\operatorname{Hom}_{\bf Ring}(A,\mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $\mathcal{A}$ be a quasi-coherent $\mathcal{O}_S$-algebra; one can prove that the function
$$
\Phi:(f,f^{\sharp})\in\operatorname{Hom}_{S-\bf Sch}(X,\operatorname{Spec}\mathcal{A})\to f^{\sharp}(\operatorname{Spec}\mathcal{A})\in\operatorname{Hom}_{\mathcal{O}_S-\bf Alg}(\mathcal{A},p_{*}\mathcal{O}_X)
$$
is bijective; where $\operatorname{Spec}\mathcal{A}$ is the relative spectrum of $\mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $\operatorname{Spec}(A\otimes_{\mathbb{C}}B)$ is canonically isomorphic to $\operatorname{Spec}A\times_{\operatorname{Spec}\mathbb{C}}\operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
Best Answer
Old answer when $R,S$ were domains:
As Mohan surmises in his comment, $f$ is an isomorphism. This is because $f:R\to S$ expresses $S$ as a finite (thus integral) extension of $R$ inside $K=Frac(R)=Frac(S)$, and $R$ is integrally closed, so $R=S$. One may now quickly verify from the definition that the property you wish for holds.
New answer now that $R,S$ are not required to be domains:
Any normal noetherian ring is a finite product of domains. On the scheme side, this means that a noetherian normal scheme is a finite disjoint union of it's irreducible (and thus connected) components. By the definition of birationality, we see that there's a unique component of $S$ over each component of $R$, so after base change along the inclusion of this component (and the fact that finite morphisms are preserved under base change) we see we're back at the previous situation and $f$ must also be an isomorphism.