YES. 
Consider the Jolissaint--Lafforgue Sobolev algebra $H_\ell^s(\Gamma)$. 
(I don't know the common name for it.)
Here we take $\Gamma=F_\infty$ to be the free group of countably infinite rank, $\ell$ the standard word length, and $s>2$. 
It is the completion of the complex group algebra ${\mathbb C}\Gamma$ under 
the Sobolev norm
$$\| f \|= (\sum_ x |f(x)|^2(1+\ell(x))^{2s})^{1/2}.$$
V. Lafforgue (https://mathscinet.ams.org/mathscinet-getitem?mr=1774859) has proved that 
$H_\ell^s(\Gamma)$ is a "Banach algebra" (see the comment below) which is embedded densely in the reduced group $\mathrm{C}^*$-algebra  $\mathrm{C}^*_\lambda(\Gamma)$ and is closed under the holomorphic functional calculus there. 
From the latter property, we see that $H_\ell^s(\Gamma)$ is simple, because the C*-algebra 
$\mathrm{C}^*_\lambda(F_\infty)$ is simple.
The Banach algebra $H_\ell^s(\Gamma)$ is unital and isomorphic to a Hilbert space as a Banach space.
For the standard free basis $\{s_n\}$ of $\Gamma=F_\infty$, the corresponding "unitary" 
elements are uniformly bounded in $H_\ell^s(\Gamma)$. Property (3) follows from this.

Comment: Note that the above Sobolev norm only satisfies $\|f * g\|\le C\|f\|\|g\|$ for some universal constant $C$, 
but one can renorm it via $H_\ell^s(\Gamma)\hookrightarrow B(H_\ell^s(\Gamma))$ to make 
it satisfies $\|f * g\|'\le \|f\|'\|g\|'$ and $\|1\|'=1$. Note that by Lumer's theorem, a unital infinite-dimensional Banach algebra cannot be isometric to a Hilbert space.