In general, no: for certain $A$ one can get non-zero $c\in A$ and sequences $(a_n)$ and $(b_n)$ in $A$ that converge weakly to zero, such that $a_n b_n=c$ for all $n$; these will show that $\ker\sigma$ is not even closed under multiplication, let alone an ideal.

We can arrange for $A$ to be Arens regular, so that $A^{**}$ is even a dual Banach algebra.

(As a loose analogy: one should suspect that the question has a negative answer just because $\lim_i \lim_j a_{ij}$ can be very different from $\lim_n a_{nn}$ for a doubly-indexed sequence of scalars.)

Here is one example, which I probably learned from conversations with Matt Daws; there should be many others with similar behaviour. I use $\ell_2$ to denote $\ell_2({\bf N})$.
Take $A= K(\ell_2)$, let $a_n$ be the operator that sends $\delta_n$ to $\delta_1$ and all other basis vectors to $0$; and let $b_n$ be the operator that sends $\delta_1$ to $\delta_n$ and all other basis vectors to $0$.
Clearly $a_nb_n$ is the projection onto the span of $\delta_1$, for all $n$.

It only remains to justify the claim that $a_n \stackrel{w}{\longrightarrow} 0$ and $b_n \stackrel{w}{\longrightarrow} 0$. By trace duality every functional on $A$ is of the form $x\mapsto {\rm Tr}(yx)$ for some trace class operator $y$. A direct calculation shows that ${\rm Tr}(ya_n) = \langle y\delta_1, \delta_n\rangle\to 0$ as $n\to\infty$, and likewise ${\rm Tr}(yb_n)=\langle y\delta_n,\delta_1 \rangle = \langle \delta_n, y^*\delta_1\rangle \to 0$.