Why is the fundamental group not a topological group? It is widely known that group operations in $\Omega X$ are generally not consistent with the natural topology (see, for example, https://arxiv.org/pdf/1105.6363.pdf). Moreover, if the operations of the product of loops and the reverse loop were always continuous until the identification of homotopic loops, then they would remain since a quotient map is open. This means that already in the algebraic structure $\langle \mathrm{Hom}(S^1, X), +, -\rangle$ (where $+$ is the concantation of loops, and $-$ is the inverse loop), operations are generally not consistent with the natural (compact-open) topology.
Then there should be an error in the proof below, but I don't see it.
Stat. The concantation in $\mathrm{Hom}(S^1, X)$ is continuous. 
Proof. Take any element of a prebase $U_K = \{f \colon S^1 \to X ~|~ f(K)\subset U\}$. Let's show that its preimage is open. Its preimage is the set of all pairs of loops (a, b) such that $a(K_1) \subset U$ and $b(K_2) \subset U$, where $K_1 = 2(K \cap [0, 1/2])$ and $K_2 = 2(K \cap [1/2, 1]) - 1$. Thus, the preimage is the direct product $U_{K_1} \times U_{K_2}$, which is open by the definition of the direct product topology
Stat. The inversion in $\mathrm{Hom}(S^1, X)$ is continuous.
Proof. The preimage of a prebase element $U_K$ is $U_{-K}$.
 A: Your arguments are correct, and the operations on $\operatorname{Hom}(S^1,X)$ are continuous.  However, this does not imply the corresponding operations on the quotient $\pi_1(X)$ are continuous.  In general, the quotient map $p:\operatorname{Hom}(S^1,X)\to \pi_1(X)$ is not open.  For instance, if $X$ is the Hawaiian earring, let $U=X\setminus \{x\}$ where $x$ is a point in the first of the circles and let $V\subseteq\operatorname{Hom}(S^1,X)$ be the open set of loops whose images are contained in $U$.  If $p(V)$ were open, that would mean $p^{-1}(p(V))$, the set of loops that are homotopic to a loop that avoids $x$, is open.  But now consider the commutator loop $c_n=a_1a_na_1^{-1}a_n^{-1}$, where $a_n$ is the loop that goes around the $n$th circle of $X$ once.  As $n\to\infty$, these loops $a_n$ converge to just the constant loop $e$ at the basepoint, so $c_n$ converges to the loop $c=a_1ea_1^{-1}e$ which is nullhomotopic and thus in $p^{-1}(p(V))$.  However, none of the $c_n$ are in $p^{-1}(p(V))$ (if they were, then after composing with the retraction $X\to S^1\vee S^1$ onto just the first and $n$th circles you would conclude that the commutator of the two generators of the free group $\pi_1(S^1\vee S^1)$ is a power of one of the generators).  This shows that $p^{-1}(p(V))$ is not open.
