I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:
Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.
Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.
Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.
Something like this ought to work for $A_5$.
EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.
Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.
The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.
Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.
Two distinct $7$-Sylow subgroups intersect at the identity element (think about Langrange's Theorem). Hence, if there are $15$ distinct $7$-Sylow subgroups, it follows that, in addition to the identity element, there are at least $15\cdot 6 = 90$ elements of order $7$. But then we only have $15=105-90$ elements to account for. You can now argue that either $n_{3}=1$ or $n_{5}=1$.
Now, whenever $n_{p}=1$, this means there exists exactly one $p$-Sylow subgroup. Using Sylow's Second Theorem, this unique $p$-Sylow subgroup must be normal. As a result, the underlying group cannot be simple.
EDIT: As you-sir-33433 remarks, the normality of $p$-Sylow subgroup simply follows from the fact that $H$ and $gHg^{-1}$ have same cardinality for a subgroup $H\subset G$, and for any $g\in G$. One does not need Sylow's second theorem for this.
Best Answer
In the hope that this will be the definitive answer to understanding groups of order $105$, I will talk about the ways to solve this.
The question assumes that the Sylow $3$-subgroup is normal in $G$. The condition on the Sylow $3$-subgroup here is necessary. There are two groups of order $105$, both with a normal Sylow $5$- and $7$-subgroup, but one is cyclic and the other is $C_5\times F_{21}$, where $F_{21}$ is a non-abelian group, the normalizer of a Sylow $7$-subgroup of $A_7$.
The fastest way to proceed is to notice that $P_3$, the Sylow $3$-subgroup, is not only normal but central. To see this, you can recall that $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$, which has order $2$ in this case. The from scratch method is to notice that $C_3$ has only two non-identity elements, so for any element $g\in G$, $g^2$ must act trivially on $P_3$. But $|G|$ is odd, so every element is a square, and $P_3$ is central.
At this point, there are two ways to proceed. The first is to notice that $G/P_3$ has order $35=5\times 7$, and groups of order $35$ are cyclic. If $G/Z(G)$ is cyclic then $G$ is abelian, and we are done. (Clearly $G$ is therefore in fact cyclic.)
The alternative proof is to note that $P_3\leq C_G(P_5)$ and $P_3\leq C_G(P_7)$. Thus $|C_G(P_5)|\geq 15$, and $|C_G(P_7)|\geq 21$. (Recall that $C_G(P_q)\leq N_G(P_q)$ and $n_q$, the number of Sylow $q$-subgroups, is equal to $|G:N_G(P_q)|$.) From Sylow's theorem ($n_q\equiv 1\bmod q$) we see that $n_5=n_7=1$, as needed.
If you don't want to do this you can count elements, although it's more subtle than most such arguments. Let's do this without the assumption that $n_3=1$, to obtain the full classification.
The number $n_5$ of Sylow $5$-subgroups is either $1$ or $21=3\times 7$. We want to prove the former, so assume the latter. Then there are $21\times 4=82$ elements of order $5$, and since $C_G(P_5)=P_5$, there are no elements of order $5n$ for any $n>1$. This leaves exactly $105-82=23$ elements of order not $5$, and these must have order $1$, $3$, $7$ or $21$. If $n_7\neq 1$ then $n_7=15$, but this is impossible as there are only $23$ elements left. So $n_7=1$, removing six elements of order $7$. There are seventeen elements left, so $n_3\leq 8$ (as each Sylow $3$-subgroup requires two elements of order $3$). Thus $n_3=1$ or $n_3=7$. If $n_3=7$ then that removes fourteen elements of order $3$, and the identity, so there are two elements remaining, which must have order $21$. But in any cyclic group of order $21$ there are twelve elements of order $21$, which is too many.
Thus $n_3=1$, and the Sylow $3$- and $7$-subgroups are both normal. Thus $P_3P_7$ is normal in $G$, has index $5$, and therefore contains every element of order dividing $21$. So where are the two remaining elements? This yields a contradiction, so $n_5=1$.
If $n_7\neq 1$ then $n_7=15$, as it must be $1$ modulo $7$. Again, you can obtain a contradiction as before, because $C_G(P_5)$ contains $P_7$ but $C_G(P_7)$ does not contain $P_5$. Let's try to count elements, and see what goes wrong. This yields $15\times 6=90$ elements of order $7$. There are five elements in $P_5$, leaving ten elements. Thus $n_3\leq 5$, so $n_3=1$. Thus we have a subgroup $P_3P_5$ of order $15$. This contains ten more elements (as we have already counted $P_5$), and so we have exactly the right number of elements, $105$.
If $15$ were a prime, then this would be fine. Then $7\mid (15-1)$ and there would be a map from $C_7$ into $\mathrm{Aut}(C_{15})$, which would have order $14$. But $15$ is not a prime, so we can obtain a contradiction using centralizers, as above, but element counting will not work in this case. The group $P_3P_5$ has normal subgroups $P_3$ and $P_5$, on which $P_7$ cannot act. Thus $P_3P_5$ is actually central, and $G/(P_3P_5)$ is cyclic, so $G$ is abelian. Alternatively, $P_3$ is central, so $P_3$ centralizes $P_7$. But $n_7=15$, so $P_7$ does not centralize $P_3$. This is a clear contradiction.
Thus $n_7=1$ as well. The subgroup $P_5P_7$ is a normal, cyclic, subgroup of order $35$. Since there is no map from $P_3$ to $\mathrm{Aut}(P_5)$, this is actually central. The subgroup $P_7P_3$, of order $21$, complements this, so $G\cong P_5\times P_7P_3$. If $n_3=1$, equivalently $P_3$ centralizes $P_7$, then you end up with an abelian (cyclic) group of order $21$. If $n_3=7$, equivalently $P_3$ acts non-trivially on $P_7$, then $P_3P_7$ is a Frobenius group of order $21$. This is the normalizer in $A_7$ of a Sylow $7$-subgroup.