Consider the space $L^1(\mathbb{R})$ which is a Banach-algebra when taking the convolution as the algebra product and even posses the B*-property if one takes $f^*(t)=\overline{f(-t)}$ as the involution. But $L^1(\mathbb{R})$ does not have the C* property.
What would be a counter example to show that $L^1(\mathbb{R})$ does not have the C* property?
Take for example any odd, continuous and real-valued $f\in L^1(\mathbb R)\cap L^2(\mathbb R)$ which is positive on $(0,\infty)$. (The requirement $f\in L^2$ ensures that $f*f^*(x)$ is defined everywhere). Then $f^*=-f$, and so $$ \vert f*f^*(x)\vert=\left\vert \int_{\mathbb R} f(t)f(x-t)\, dt\right\vert \leq \int_{\mathbb R} \vert f(t)\vert \,\vert f(x-t)\vert \, dt $$ for all $x\in\mathbb R$. Moreover, we have $strict$ inequality for any $x> 0$ because the function $t\mapsto f(t)f(x-t)$ changes sign at $t=x$. It follows that $\Vert f*f^*\Vert_{L^1}=\int_{\mathbb R} \vert f*f^*(x)\vert dx$ is $strictly$ smaller than $\int_{\mathbb R}\int_{\mathbb R} \vert f(t)\vert \,\vert f(x-t)\vert \, dt dx=\Vert f\Vert_{L^1}^2$