EDIT: The definition literally says that the product is bilinear, as Chris Eagle kindly pointed out. Turns out that my original proof was fine. This question does no longer require an answer and therefore can be closed (if that's appropriate). Thank you.
In the book "Functional Analysis", by Bühler and Salamon, there is the following definition of a Real Banach Algebra:
A real Banach Algebra is a pair consisting of a real Banach space $(\mathcal{A},\|\cdot \|)$ and a bilinear map $\mathcal{A}\times \mathcal{A}\to \mathcal{A}:(a,b)\mapsto ab$ (called the product) that is associative, i.e.$$(ab)c=a(bc)\qquad \forall a,b,c\in \mathcal{A}$$and satisfies the inequality$$\|ab\|\leq \|a\|\|b\|\qquad \forall a,b\in \mathcal{A}.$$
I know that the property $\|ab\|\leq \|a\|\|b\|$ for all $a,b\in \mathcal{A}$ is also called continuity of product, so I wanted to prove that this in fact implies that for a fixed $b$, the map $\mathcal{A}\to \mathcal{A}:a\mapsto ab$ is continuous. If the multiplication by a fixed number were linear, then I could set a sequence $(a_n)$ converging to $a$ and use that$$0\leq \|a_nb-ab\|=\|(a_n-a)b\|\leq \|a_n-a\|\|b\|\xrightarrow{n\to \infty}0$$to obtain that $a_nb\xrightarrow{n\to \infty}ab$, and therefore the product would indeed be continous.
However, whitout the linearity of the product I can only go as far as$$0\leq \|a_nb-ab\|\leq \|a_nb\|+\|ab\|\leq \|a_n\|\|b\|+\|a\|\|b\|=(\|a_n\|+\|a\|)\|b\|$$but this doesn't imply that $\|a_nb-ab\|\xrightarrow{n\to \infty}0$. Is there another way of proving that the product is continuous?
