The answer is **no** (a unit is not necessary within the algebra).
In the case the product is continuous (i.e. there exists $M>0$ such that, identically, $||xy||\leq M||x||.||y||$ see this question) and of a unital algebra, the common way to construct a multiplicative norm equivalent to the given one is to consider the left-regular representation i.e. $s\rightarrow \gamma(s)$ where $\gamma(s)\in \mathcal{L}_A$ is defined by $\gamma(s)[x]=sx$ and setting the new norm $||s||':=|||\gamma(s)|||$ (the last norm being that of the *bounded convergence* within $\mathcal{L}_A$). In this respect, the representation $s\rightarrow \gamma(s)$ must be *faithful* (it is the case, in particular, when $A$ is unital but not only). If it is not, then $||\ ||_1$ is only a seminorm. Following Yemon's comment, in the case when the given algebra is not unital, a way to circumvent this is to consider the left-regular $s\rightarrow \gamma_1(s)$ representation of $A$ on the Banach space $A\oplus\mathbb{C}$ (endowed, for example, with the norm $||x+\lambda||_1=||x||+|\lambda|$). This representation is given by $\gamma_1(s)[x+\lambda]=sx+\lambda.s$. It is an easy exercise to prove that the new seminorm $||s||'':=|||\gamma_1(s)|||$ is, in any case, a norm equivalent to the given one. Hope that it helps.

in which multiplication (right and left) is continuous? $\endgroup$faithful. If your Banach algebra has no zero divisor the statement works (with onlyonemultiplication). $\endgroup$