As you observed correctly the symmetric $2 \times 2$ matrices are not a ring (with the usual operations) since the set is not closed under multiplications; it is at least an additive subgroup though.
Examples where already given, but let me add one.
$$\begin{pmatrix}
a & b \\
b & d
\end{pmatrix} \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} = \begin{pmatrix}
0 & b \\
0 & d
\end{pmatrix} $$
which is not symmetric (unless $b=0$).
It is however true that the matrices of the form
$$\begin{pmatrix}
a & b \\
b & a
\end{pmatrix}
$$
form a ring and possibly this was what was meant in your source.
Indeed, the matrices form clearly an additive subgroup, and writing $I$ for the identity matrix and $J$ for the matrix $$\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
$$
each such matrix can be written (uniquely) as $xI+ yJ$ with real $x,y$. Since we have $I^2 = I$, $IJ=JI= J$ and $J^2 = I$, and the distributive law, it follows that the matrix is again of the claimed form.
Furthermore, we get from this consideration that this ring is isomorphic to $\mathbb{R}[X]/(X^2 -1)$. Since $X^2 - 1$ is not irreducible over $\mathbb{R}$ this is not a field, and we get that a further isomorphism to $\mathbb{R}[X]/(X -1) \times \mathbb{R}[X]/(X +1) \cong \mathbb{R} \times \mathbb{R}$.
In this context it might also be interesting to recall that one way to construct the complex numbers is as matrices of the form
$$\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
$$
There the polynomial we get is $X^2 + 1$, which is irreducible over $\mathbb{R}$.
