Just to contribute a little, let me give a von Neumann algebra perspective.

Let $B=A'$, i.e. the commutant of $A$ in $G$. Then 2) is immediately satisfied, and so is 3), because $A\cap B$ is the centre of $A$, which is trivial.

As we are in a type I factor, $B'=A$.

Also \[ \alg(A,B)=(A\cup B)''=(A'\cap B')'=(A'\cap A)'=(\mathbb{C}\,I)'=G, \] which shows that $G$ is the algebra generated by $A$ and $B$. It remains to see that $G$ can be identified with $A\otimes B$. The map $a\times b\mapsto ab$, extended by linearity, is a *-homomorphism from $A\otimes B$ to $G$. It is clearly onto. And it is also one-to-one, because if $\sum_ja_jb_j=0$, one can use the idea in 11.1.8 in Kadison-Ringrose to see that $\sum_ja_j\otimesb_j=0$ (it is essential the fact that $a_jb_j=b_ja_j$). Finally, a bijective *-homomorphism is also isometric.

Just to contribute a little, let me give a von Neumann algebra perspective.

Let $B=A'$, i.e. the commutant of $A$ in $G$. Then 2) is immediately satisfied, and so is 3), because $A\cap B$ is the centre of $A$, which is trivial.

As we are in a type I factor, $B'=A$.

Also $$ \text{alg}(A,B)=(A\cup B)''=(A'\cap B')'=(A'\cap A)'=(\mathbb{C}\,I)'=G, $$ which shows that $G$ is the algebra generated by $A$ and $B$. It remains to see that $G$ can be identified with $A\otimes B$. The map $a\times b\mapsto ab$, extended by linearity, is a *-homomorphism from $A\otimes B$ to $G$. It is clearly onto. And it is also one-to-one, because if $\sum_ja_jb_j=0$, one can use the idea in 11.1.8 in Kadison-Ringrose to see that $\sum_ja_j\otimes b_j=0$ (it is essential the fact that $a_jb_j=b_ja_j$). Finally, a bijective *-homomorphism is also isometric.