If I'm not wrong, the following statement is true :
Theorem. Let $E$ be a (finite or infinite) set, let $G$ be a finite subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Then all the non-identity elements of $G$ have the same fixed point. (And thus, if $G$ is not trivial, there is one and only one element $x$ of $E$ that is fixed by every element of $G$. Then $G$ acts freely on $E \setminus \{x\}$.)
I looked for a proof in textbooks and on Internet, but I saw nothing. (Maybe I looked badly.) I found a proof myself (I sketch it below) but it is not very beautiful and I am afraid it is too complicated. Thus my question is : do you know a more direct proof ?
Here is my proof.
$\mathbf{Step 1.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that $G$ is abelian. Then all the non-identity elements of $G$ have the same fixed point.
$\mathbf{Proof.}$ Let $\alpha$ and $\beta$ non-identity elements of $G$. Since $G$ is abelian,
$\alpha ^{-1} \beta \alpha = \beta$.
Applying both members to the unique fixed point $b$ of $\beta$ gives
$\alpha ^{-1} \beta \alpha (b)= b$.
Applying $\alpha$ to both members gives
$\beta \alpha (b) = \alpha (b)$, thus $\alpha (b)$ is a fixed point of $\beta$. Since $b$ is the only fixed point of $\beta$, we have thus $\alpha (b) = b$, thus $b$ is the fixed point of $\alpha)$, thus $\alpha$ and $\beta$ have the same fixed point. This proves step 1.
$\mathbf{Step. 2.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that $G$ has a nontrivial normal subgroup whose all non-identity elements have the same fixed point. Then all the non-identity elements of $G$ have the same fixed point.
$\mathbf{Proof.}$ By hypothesis, we can choose a nontrivial normal subgroup $H$ of $G$ such that all the non-identity elements of $H$ have the same fixed point.
Choose a non-identity element $\alpha$ of $H$. From the hypotheses,
(1) $\alpha$ has a unique fixed point, say $a$, and
(2) every non-identy element of $H$ has $a$ as unique fixed point.
Let $\gamma$ be a non-identity element of $G$. Since $H$ is normal in $G$, $\gamma ^{-1} \alpha \gamma$ is a non-identity element of $H$, thus, in view of (2),
$\gamma ^{-1} \alpha \gamma (a) = a$. Applying $\gamma$ to both members gives $\alpha \gamma (a) = \gamma (a)$, thus $\gamma (a)$ is a fixed point of $\alpha$. Thus, by (1), $\gamma (a) = a$. By hypothesis, $\gamma$ has only one fixed point, thus our result means that the only fixed point of $\gamma$ is $a$. This is proven for every non-identity element $\gamma$ of $G$, thus step 2 is proven.
$\mathbf{Step. 3.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that there is a generating subset $X$ of $G$ such that all the non-identity elements of $X$ have the same fixed point. Then all the non-identity elements of $G$ have the same fixed point.
$\mathbf{Proof.}$ It is an easy consequence of the fact that every element of $G$ is a product of non-identity elements of $X \cup X^{-1}$.
$\mathbf{Step. 4.}$ Let $E$ be a (finite or infinite) set, let $G$ be a (finite or infinite) subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Assume that there are two differet maximal subgroups $M_{1}$ and $M_{2}$ such that
(i) $M_{1} \cap M_{2} \not= 1$,
(ii) all the non-identity elements of $M_{1}$ have the same fixed point and
(iii) all the non-identity elements of $M_{2}$ have the same fixed point.
Then all the non-identity elements of $G$ have the same fixed point.
$\mathbf{Proof.}$ From hypotheses (i), (ii) and (iii), it results that
(1) all the non-identity elements of $M_{1} \cup M_{2}$ have the same fixed point.
On the other hand, since $M_{1}$ and $M_{2}$ are two different maximal subgroups of $G$, they generate $G$, in other words,
(2) $M_{1} \cup M_{2}$ is a generating subset of $G$.
By (1), (2) and step 3, all the non-identity elements of $G$ have the same fixed point, thus step 4 is proven.
$\mathbf{Step. 5.}$ Let $E$ be a (finite or infinite) set, let $G$ be a $\mathbf{finite}$ subgroup of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point. Then all the non-identity elements of $G$ have the same fixed point.
$\mathbf{Proof.}$ Assume, by contradiction, that
(hyp. 1) the statement is false.
Thus, there exist a set $E$ and a finite subgroup $G$ of $S_{E}$ such that every non-identity element of $G$ has exactly one fixed point and the non-identity elements of $G$ do not all have the same fixed point. Among these subgroups $G$ of $E$, choose $G_{0}$ with the least possible order. Then
(2) $G_{0}$ is a finite subgroup of $S_{E}$,
(3) every non-identity element of $G_{0}$ has a unique fixed point,
(4) the non-identity elements of $G_{0}$ do not all have the same fixed point,
and, in view of the minimality of $\vert G_{0} \vert$,
(5) for every proper subgroup $K$ of $G_{0}$, all non-identity elements of $K$ have the same fixed point.
In view of (3), (4), (5) and step 4,
(6) the maximal subgroups of $G_{0}$ intersect pairwise trivially.
Assume that
(hyp. 7) $G_{0}$ has a normal subgroup $H$ such that $1 < H < G_{0}$.
By (5) (and the assumption $H < G_{0}$), all non-identity elements of $H$ have the same fixed point. Thus, by step 2 (and the assumption $1 < H$), all non-identity elements of $G_{0}$ have the same fixed point. This contradicts (4), thus (hyp. 7) is absurd, thus $G_{0}$ is a simple group. Thus, by (2) and step 1,
(8) $G_{0}$ is a finite non-abelian simple group.
Now, (6) and (8) are incompatible, as proven here :
Thus our hypothesis (1) is absurd, so step 5 is proven.
Best Answer
If you are familiar with the theory of finite Frobenius groups, then it might be easier to use that. The action of the group on any non-regular orbit of length greater than $1$ must be as a Frobenius group, and it must be a faithful action. But then the fixed point free elements in the action cannot fix any points, because they cannot also lie in a Frobenius complement.
Note that the result is false for infinite groups, because there are infinite Frobenius groups in which all elements fix a unique point. (See, for example, Timm von Puttkamer's answer here.)