In *general* Banach algebras the spectral radius is neither subadditive nor submultiplicative; in particular, neither of the two properties you mention holds.


$2\times 2$ (real or complex) matrices should suffice to give examples, so this is not to do with any subtleties of infinite-dimensional algebras.

For example, take $ a= \left(\begin{matrix} 0 & 1 \\\\ 0 & 0 \end{matrix} \right) $.
Note that this is nilpotent, so the only point in the spectrum is zero, and hence $\sigma(a)\sigma(b)= \{0\}$ for any other matrix $b$. On the other hand, we can find $b$ for which $ab$ is not nilpotent, so that $\sigma(ab)\not\subseteq \{0\}$. A simple choice which works is $b=\left(\begin{matrix} 0 & 0 \\\\ 1 & 0 \end{matrix} \right)$, since then $ab$ is a non-zero projection (=idempotent) and so contains $1$ in its spectrum.

The same pair also works as a counter-example for the "additive question". For since $\sigma(a)=\sigma(b)=\{0\}$, we have $\sigma(a)+\sigma(b)=\{0\}$. On the other hand, $a+b$ is a reflection and hence its spectrum is $\{-1,1\}$