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 $\mathrm{C}^*$-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 and their inverses 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.

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. (Property RD for $F_\infty$ is due to Haagerup.) From the latter property, we see that $H_\ell^s(\Gamma)$ is simple, because the $\mathrm{C}^*$-algebra $\mathrm{C}^*_\lambda(F_\infty)$ is simple (Powers). 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 and their inverses 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.

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.

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 $\mathrm{C}^*$-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 and their inverses 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.