If $G$ is a finite group in which the centralizer of every nonidentity element is cyclic, then as Geoff Robinson pointed out, the Fitting subgroup $F = F(G)$ is cyclic, and $G/F$ is cyclic. One can say more, however. Suppose that $G$ is not itself cyclic, so $F < G$. Let $x$ be an element whose image modulo $F$ generates $G/F$, so $x \ne 1$. Let $C = C_G(x)$, so $C$ is cyclic, and since $x \in C$, we have $FC = G$. Since both $F$ and $C$ are cyclic, their intersection centralizes both $F$ and $C$, so is central in $G$. If this intersection contains a nontrivial element $y$, then $G =C_G(y)$, so $G$ is cyclic, contrary to assumption. Thus $F \cap C = 1$, and so $G$ can be constructed as a semidirect product of $F$ acted on by $C$.
One can say still more. Suppose some nonidentity element $c \in C$ commutes with some nonidentity element $f \in F$. Then $C_G(f)$ contains $F$ and also contains $c$. Since $c \not\in F$, we see that $C_G(f)$ strictly contains $F$. This is impossible, however, because $C_G(f)$ is cyclic, and yet $F$ is its own centralizer in $G$. This shows that the action of $C$ on $F$ is "Frobenius", and in particular, $|C|$ divides $|F|-1$ so $|F|$ and $|C|$ are coprime.
Conversely, if we start with an arbitrary finite cyclic group $F$, we can build all possible finite groups $G$ satisfying the cyclic-centralizer assumption and such that $F(G) = F$. Do this as follows. Let $d$ be the GCD of all of the numbers $p-1$ for primes $p$ dividing $|F|$, and let $C$ be any cyclic group of order dividing $d$. Then $C$ has a unique Frobenius action on $F$ and the semidirect product $G$ of $F$ by $C$ will have the desired property. The key to checking that $G$ does have this property is that every element of $G$ either lies in $F$ or is conjugate to an element of $C$, so it is enough to check that centralizers of nonidentity elements of $F$ and of $C$ are cyclic. These centralizers are respectively $F$ and $C$.