I would avoid the counting argument, since there are indeed groups of order $60$ with $15$ Sylow $2$-subgroups, and all elements of order $2$ are shared by several of these Sylow subgroups at once.
Here is a slightly easier way to do this, using a slightly stronger version of Sylow's theorems. Namely,
Theorem: Let $p$ be a prime, and $p^r$ the highest power of $p$ dividing $G$. For any two Sylow $p$-subgroups $P$ and $Q$ of $G$, suppose we have $p^k\le |P\cap Q|$. Then the number of Sylow $p$-subgroups of $G$ - call it $n_p$ - satisfies $n_p\equiv 1\pmod{p^{r-k}}$.
The proof is exactly the same as the usual proof (using group actions), you just have to pay extra attention to orbit sizes along the way.
Now, if $G$ is a group of order $60$, containing $15$ Sylow $2$-subgroups, then since $15\not\equiv 1\pmod{4}$, the theorem implies the existence of two Sylow $2$-subgroups - call them $P$ and $Q$ - with $|P\cap Q|=2$. But then $P\cap Q$ is normal in both $P$ and $Q$, and thus its normalizer, $N_G(P\cap Q)$, has order divisible by $4$, and size at least $|P|+|Q|-|P\cap Q|=6$. Thus $N_G(P\cap Q)$ has size at least $12$.
The action of $G$ on the right cosets of $N_G(P\cap Q)$ gives a homomorphism from $G$ into $S_5$. If we are assuming $G$ is simple, this becomes an isomorphism with $A_5$.
[Note that this is actually a contradiction, since $A_5$ does not have $15$ Sylow $2$-subgroups, but either way, you are done.]
As requested, here is a summary of the points raised in the comments.
The argument as posted in the question is very nearly complete.
As is true for any homomorphism, $Ker(\phi)$ is a normal subgroup. Of course, $S_5$ really hasn't got any normal subgroups. There is $S_5$ itself, the trivial group, and $A_5$. Now, as was sorted out in the question itself, the fact that all the $5$-Sylow subgroups are conjugate implies that $Im(\phi)$ is non-trivial, it has to have at least $6$ elements! It follows that $Ker(\phi)≤20$. That rules out both the complete group $S_5$ and the alternating group $A_5$, so $Ker(\phi)$ must be trivial. As was pointed out (correctly) in the question this is all that was needed to complete the argument.
Best Answer
Let $|G|=2^n \cdot 3, \ \ n \geqslant 2$ and $H$ Sylow 2-subgroup of $G$. Then $|G:H|=3$, so $|G:\operatorname{Core}(H)| | 3!=6$. So, $\operatorname{Core}(H)$ is non-trivial group, because $4| |G|$ and $4 \nmid6.$ $\operatorname{Core}(H)$ is normal, and she is not whole group because $\operatorname{Core}(H) \le H \lneq G$. So, $G$ is not simple group.