The untwisted Hall product is associative and unital

Fix basis objects $M,N,P,L$. The coefficient of $[L]$ in $([M]\ast [N])\ast [P]$ is the sum over isomorphism classes $[X]$ of $g_{M,N}^X{g}_{X,P}^L$. By [F1], this counts a subobject $W\le L$ with $W\cong P$ and $L\mathrm{/}W\cong X$, together with a subobject $\bar{U}\le L\mathrm{/}W$ with $\bar{U}\cong N$ and $(L\mathrm{/}W)\mathrm{/}\bar{U}\cong M$. The quotient class $[X]=[L\mathrm{/}W]$ is uniquely determined by $W$, so this sum counts exactly these pairs.

By [F3], each such pair $(W,\bar{U})$ corresponds to exactly one flag $W\le U\le L$ with $W\cong P$, $U\mathrm{/}W\cong N$, and $L\mathrm{/}U\cong M$. The coefficient of $[L]$ in $[M]\ast ([N]\ast [P])$ counts the same flags: first choose $U\le L$ with $U\cong Y$, $L\mathrm{/}U\cong M$, then choose $W\le U$ with $W\cong P$, $U\mathrm{/}W\cong N$, summing over the uniquely determined class $[Y]=[U]$.

The only subobject isomorphic to zero is the zero subobject, and the only subobject with zero quotient is the whole object: in the latter case the inclusion has zero cokernel and hence is an isomorphism by [F4]. Thus $g_{M,0}^L=1$ exactly when $L\cong M$, while $g_{0,M}^L=1$ exactly when $L\cong M$. The zero-object symbol is a two-sided unit.

The two parenthesizations therefore have the same coefficient at every basis vector $[L]$, so $([M]\ast [N])\ast [P]=[M]\ast ([N]\ast [P])$ for basis elements.

Basis products have finite support by [F2]. Every element of $H_{\mathcal{A},R}$ is a finite linear combination of basis vectors, so bilinearity extends the basis calculation in step 2.1 to all triples and preserves finite support. Step 1.3 supplies the unit, proving the statement.