This is only a partial answer.

I will show that (1) and (2) are equivalent. Since the left regular representation is isometric, I will forget about $L$ and I will simply denote the operator norm by $\|a\|$. It suffices to show that if an element satisfies $\|a\|=\|a\|_{\phi}=1$ then it is unitary. In order to prove that, note that $\|a\|=1$ implies that $\|a^{\ast}a\|=1$, hence $a^{\ast}a \leqslant \text{Id}$. By positivity of the trace, we get $\phi(\text{Id}-a^{\ast}a)\geqslant 0$. On the other hand, it is equal to $0$, hence by faithfulness we obtain $a^{\ast}a=\text{Id}$. As $\phi$ is a trace, this works equally well for $aa^{\ast}$, so $aa^{\ast}=a^{\ast}a=\text{Id}$.

I don't have yet a clear idea, where to look for possible counterexamples to other statements, but maybe the following paper would help: Adrian Ioana, Sorin Popa, and Stefaan Vaes, A class of superrigid group von Neumann algebras, Ann. of Math. (2), vol. 178 (2013), no. 1, 231--286. The authors show, among other things, that corners (i.e. algebras of the form $pMp$, where $p\in M$ is a projection) of certain group von Neumann algebras are never group von Neumann algebras. It is conceivable that these corners should satisfy condition (3).

This is only a partial answer.

I will show that (1) and (2) are equivalent. Since the left regular representation is isometric, I will forget about $L$ and I will simply denote the operator norm by $\|a\|$. It suffices to show that if an element satisfies $\|a\|=\|a\|_{\phi}=1$ then it is unitary. In order to prove that, note that $\|a\|=1$ implies that $\|a^{\ast}a\|=1$, hence $a^{\ast}a \leqslant \text{Id}$. By positivity of the trace, we get $\phi(\text{Id}-a^{\ast}a)\geqslant 0$. On the other hand, it is equal to $0$, hence by faithfulness we obtain $a^{\ast}a=\text{Id}$. As $\phi$ is a trace, this works equally well for $aa^{\ast}$, so $aa^{\ast}=a^{\ast}a=\text{Id}$.

I don't have yet a clear idea, where to look for possible counterexamples to other statements, but maybe the following paper would help: Adrian Ioana, Sorin Popa, and Stefaan Vaes, A class of superrigid group von Neumann algebras, Ann. of Math. (2), vol. 178 (2013), no. 1, 231--286; doi: 10.4007/annals.2013.178.1.4, arXiv:1007.1412. The authors show, among other things, that corners (i.e. algebras of the form $pMp$, where $p\in M$ is a projection) of certain group von Neumann algebras are never group von Neumann algebras. It is conceivable that these corners should satisfy condition (3).

This is only a partial answer.

I will show that (1) and (2) are equivalent. Since the left regular representation is isometric, I will forget about $L$ and I will simply denote the operator norm by $\|a\|$. It suffices to show that if an element satisfies $\|a\|=\|a\|_{\phi}=1$ then it is unitary. In order to prove that, note that $\|a\|=1$ implies that $\|a^{\ast}a\|=1$, hence $a^{\ast}a \leqslant \text{Id}$. By positivity of the trace, we get $\phi(\text{Id}-a^{\ast}a)\geqslant 0$. On the other hand, it is equal to $0$, hence by faithfulness we obtain $a^{\ast}a=\text{Id}$. As $\phi$ is a trace, this works equally well for $aa^{\ast}$, so $aa^{\ast}=a^{\ast}a=\text{Id}$.

I don't have yet a clear idea, where to look for possible counterexamples to other statements, but maybe the following paper would help: Adrian Ioana, Sorin Popa Sorin, and Stefaan Vaes, A class of superrigid group von Neumann algebras, Ann. of Math. (2), vol. 178 (2013), no. 1, 231--286. The authors show, among other things, that corners (i.e. algebras of the form $pMp$, where $p\in M$ is a projection) of certain group von Neumann algebras are never group von Neumann algebras. It is conceivable that these corners should satisfy condition (3).

This is only a partial answer.

I will show that (1) and (2) are equivalent. Since the left regular representation is isometric, I will forget about $L$ and I will simply denote the operator norm by $\|a\|$. It suffices to show that if an element satisfies $\|a\|=\|a\|_{\phi}=1$ then it is unitary. In order to prove that, note that $\|a\|=1$ implies that $\|a^{\ast}a\|=1$, hence $a^{\ast}a \leqslant \text{Id}$. By positivity of the trace, we get $\phi(\text{Id}-a^{\ast}a)\geqslant 0$. On the other hand, it is equal to $0$, hence by faithfulness we obtain $a^{\ast}a=\text{Id}$. As $\phi$ is a trace, this works equally well for $aa^{\ast}$, so $aa^{\ast}=a^{\ast}a=\text{Id}$.

I don't have yet a clear idea, where to look for possible counterexamples to other statements, but maybe the following paper would help: Adrian Ioana, Sorin Popa, and Stefaan Vaes, A class of superrigid group von Neumann algebras, Ann. of Math. (2), vol. 178 (2013), no. 1, 231--286. The authors show, among other things, that corners (i.e. algebras of the form $pMp$, where $p\in M$ is a projection) of certain group von Neumann algebras are never group von Neumann algebras. It is conceivable that these corners should satisfy condition (3).