= Solution
For $x,y\in B$, write $L_x(y)=xy$ and $R_y(x)=xy$. Separate <continuity> and <linearity> make both maps <bounded linear operators>. For each fixed $y$,
$$
\sup_{\|x\|\leq1}\|L_x y\|=\sup_{\|x\|\leq1}\|R_yx\|\leq\|R_y\|<\infty.
$$
The <Uniform boundedness principle> therefore gives $C<\infty$ such that $\|L_x\|\leq C$ whenever $\|x\|\leq1$. Scaling yields
$$
\|xy\|\leq C\|x\|\|y\|.
$$
For completeness, the relevant <Uniform boundedness principle> follows from the <Baire category theorem> as follows. For a pointwise bounded family $\mathcal T$ of <bounded linear operators> on a <Banach space>, the <closed sets> $E_m=\{y:\sup_{T\in\mathcal T}\|Ty\|\leq m\}$ cover the space. One $E_m$ contains an <open ball> $B(y_0,r)$. Subtracting the bounds for $y_0$ and $y_0+h$, with $\|h\|<r$, gives $\sup_T\|Th\|\leq2m$. Applying this to $h=(r/2)u$, $\|u\|\leq1$, bounds every <operator norm> by $4m/r$.
Define the new <norm> by
$$
\boxed{\|x\|_* =\|L_x\|.}
$$
It is a <norm> because $L_xe=x$, so $L_x=0$ implies $x=0$. The <triangle inequality> and scalar homogeneity follow from those of the <operator norm>. Moreover,
$$
\frac{\|x\|}{\|e\|}\leq\|L_x\|\leq C\|x\|,
$$
so it is equivalent to the original <norm> and remains complete. By <associativity>, $L_{xy}=L_xL_y$, and the <submultiplicativity> of the <operator norm> proves
$$
\boxed{\|xy\|_*\leq\|x\|_*\|y\|_*.}
$$
This is <renorming a separately continuous Banach algebra>. It even gives $\|e\|_*=1$. If the algebra is the zero space, the conclusion is immediate separately.
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