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 $\{-1,1\}$

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Jonas Meyer points out in comments that one can pose the following converse question: let $A$ be a Banach algebra with identity, and suppose that we have

($*$) $\sigma(a+b)\subseteq\sigma(a)+\sigma(b)$ and $\sigma(a)\sigma(b)\subseteq\sigma(ab)$ for all $a,b\in A$.

Must $A$ be commutative?

As Jonas also pointed out in comments, the answer is in general `no': the algebra of upper-triangular matrices (or, more precisely, the algebra of scalar+upper triangular $m\times m$ matrices, for some fixed $m$) gives a counterexample, at least when $m\geq 3$. 

A more careful version of this argument shows, I think, that if $A$ is a finite-dimensional algebra with identity, such that $A/{\rm Rad}(A)$ is commutative, then $A$ will satisfy condition $(*)$. I suspect that the same might be true for any unital Banach algebra that is commutative modulo its radical, i.e. that the finite-dimensional hypothesis is unnecessary; but at present I'm a bit too tired to check this properly.

We could therefore modify the question yet further, and ask if a Banach algebra that satisfies condition $(*)$ must be commutative modulo its radical. The answer turns out to be *yes*, after a wander down memory lane and a forage on MathSciNet:

MR0461139 (57 #1124)
J. Zem&aacute;nek. <i>Spectral radius characterizations of commutativity in Banach algebras.</i>
Studia Math. 61 (1977), no. 3, 257--268.

The MR is short and informative enough to give in full, for those without access:

<blockquote>
It is standard that the spectral radius is subadditive and submultiplicative on any commutative complex Banach algebra. The author proves that, for a complex Banach algebra $A$, the following three conditions are equivalent: (1) the spectral radius is sub-additive on $A$, (2) the spectral radius is submultiplicative on $A$, (3) $A$ is commutative modulo its radical. Some applications of this result to other problems in Banach algebras are given, along with references to a number of related papers.
</blockquote>