Proof:

Denoting the common two sided identity by $1$,

$$(f \ast g) = (f \ast^\prime 1) \ast (1 \ast^\prime g) = (f \ast 1) \ast^\prime (1 \ast g) = f \ast^\prime g$$

which proves (ii). Next we prove commutativity:

$$g \ast f = (1 \ast^\prime g) \ast (f \ast^\prime 1) = (1 \ast f) \ast^\prime (g \ast 1) = f \ast^\prime g = f \ast g.$$

Finally, using (ii) we prove associativity:

$$(f \ast g) \ast h = (f \ast g) \ast^\prime (1 \ast h) = (f \ast... ...me 1) \ast (g \ast^\prime h) = f \ast ( g \ast^\prime h) = f \ast(g \ast h)$$