Can ring homomorphisms be characterized as ring maps such that preimage of any ideal is an ideal?

No. For instance, the function from any ring to itself which reverses the sign of each element (i.e. $f(x) = -x$) will conserve ideals, but it usually won't be a homomorphism.

You can make this fail, for almost any nonzero $R$ and almost any nonzero $S$, by taking $$f(x)=\begin{cases} 0,&\ x=0\\ \ \\ s_0,&\ x\ne0\end{cases}$$ where $s_0$ is a nonzero fixed element of $S$,

Then for any ideal the pre-image will be $\{0\}$ or $R$, depending on whether the ideal contains $s_0$ or not.

In some corner cases $f$ may result in an isomorphism, as Eric mentions below.

Consider any bijection of $\mathbb R\to\mathbb R$ which isn't additive, but maps $0$ to $0$. You could, for example, take $x\neq 1$ and interchange $x$ and $1$ and map everything else to itself.

It clearly satisfies the property, since $f^{-1}(\{0\})=\{0\}$ and $f^{-1}(\mathbb R)=\mathbb R$, and those are all the ideals that there are.

This map is not a ring homomorphism though, since it does not preserve the multiplicative identity.

This trick works pretty much for any map between rings, except for some edge combinations using $\{0\}$ and $F_2$.