There is no finite set $\Omega$ on which $G \cong A_7$ acts transitive, nonregular and such that each nontrivial element has at most two fixed points. A little lemma first:
Lemma: Let $G$ act transitive and nonregular and such that each nontrivial element has at most two fixed points, and suppose $X \le G_{\alpha}$, then $|N_G(X) : N_G(X) \cap G_{\alpha}| \le 2$.
Proof: The normalizer $N_G(X)$ acts on the set of at most two fixed points of $X$ with point stabilizer $N_G(X) \cap G_{\alpha}$. $\square$
Also it is used that under the stated assumption, if a point stabilizer is divided by an odd prime $p$, then it already contains a full Sylow $p$-subgroup of $G$.
Now let $G \cong A_7$ and assume we have some finite $\Omega$ such that $G$ acts transitive, nonregular and each nontrivial element has at most two fixed points. Let $\alpha \in \Omega$ and $H := G_{\alpha} \ne 1$. We begin by showing that $H$ is a $\{2,3\}$-group. Throughout, we use facts about the subgroup structure of $A_7$.
Assume that $H$ is not a $\{2,3\}$-group and first suppose $X \le H$ is a subgroup of order $7$. As there exists a $7$-group whose normalizer is divisble by $3$, and a $3$-subgroup whose normalizer is divisible by $4$, the above Lemma yields that $H$ contains a subgroup isomorphic to $A_4$. However, $G$ does not have a proper subgroup that contains a $7$-cycle and a subgroup isomorphic to $A_4$.
Next suppose that $Q \le H$ is a subgroup of order $5$. We know that there exists a $5$-subgroup whose normalizer is divisible by $3$ and a $3$-subgroup whose normalizer is divisble by $4$, so $H$ contains a subgroup isomorphic to $A_4 \times 3$ by the above Lemma. Hence a similar argument as before applies: $G$ does not have a proper subgroup containing a $5$-cycle as well as a subgroup isomorphic to $A_4 \times 3$. Now we know that $H$ is a $\{2,3\}$-group. The above Lemma and the fact that $H$ already contains a full Sylow $3$-subgroup force that we have a $2$-subgroup whose normalizer has order divisible by $3$ and a $3$-subgroup whose normalizer has order divisble by $4$, so both primes occur in $H$ and it follows that $H$ contains a direct product of $A_4$ with a subgroup of order $3$. In particular $H$ is not a Frobenius group, and so we have $\beta \in \Omega$ such that $\beta \ne \alpha$ and $H = G_{\beta}$. There are two cases remaining:
If $|H| = 36$, then $\Omega$ has $70$ points and thus every element of order $3$ has one fixed point whereas every involution has two fixed points. This is impossible.
If $|H| = 72$, then $\Omega$ has $35$ pionts and every element of order $3$ has two fixed points whereas every involution has exactly one fixed point. Again this is impossible. $\square$
I am not sure about the application of the stated Lemma at the various points, and how it yields the conclusions; I will give more details about the points I am unsure:
1) "As there exists a $7$-group whose normalizer is divisble by $3$, and a $3$-subgroup whose normalizer is divisible by $4$, the above Lemma yields that $H$ contains a subgroup isomorphic to $A_4$."
For the $7$-group generated by $x = (1 ~ 2 ~ 3 ~ 4 ~ 5 ~ 6 ~ 7)$ we have with $y = (2~ 3 ~ 5)(4 ~ 7 ~ 6)$ that $y^{-1}xy = x^2$, hence the element $y$ of order $3$ is contained in the normalizer and obviously $(4 ~ 5 ~ 6 ~ 7)$ normalizes $z := (1 ~ 2 ~ 3)$, giving the claims about the normalizers. But how is this related to the lemma and gives the claim that $H$ contains a subgroup isomorphic to $A_4$?
2) "Next suppose that $Q \le H$ is a subgroup of order $5$. We know that there exists a $5$-subgroup whose normalizer is divisible by $3$ and a $3$-subgroup whose normalizer is divisble by $4$, so $H$ contains a subgroup isomorphic to $A_4 \times 3$ by the above Lemma."
Again, what have these relations between $p$-subgroups and their normalizers to do with the conclusions they draw with the help of the lemma? The argument seem quite obscure to me...
3) "If $|H| = 36$, then $\Omega$ has $70$ points and thus every element of order $3$ has one fixed point whereas every involution has two fixed points."
This statement, and the following one if $|H| = 72$ I do not see. I see that if $|\Omega| = 70$, then the number of fixed points of an element of order $3$ must be congruent to $1$ mod $3$, and similar the number of fixed points of every involution must be divisible by two, but why are no fixed points for the involution excluded, and the other case that we have two fixed points applies?
Hoping someone can clarify these argument for me!? Thank you!