I write $XY$ instead of $X\otimes Y$ and ignore canonical isomorphism and morphisms labels (unambiguous).

I assume the axiom:

Ax) $(AXAY\to AAXY\to AXY) = (AXAY\to XAAY\to XAY\to AXY)$.

The second member is an alternative computation of $\alpha$.

First I have to show that:

$F_A(X)F_A(Y)F_A(Z)\to F_A(XY)F_A(Z)\to F_A(XYZ) =$

$F_A(X)F_A(Y)F_A(Z)\to F_A(X)F_A(YZ)\to F_A(XYZ)$

i.e. that
$$AXAYAZ\to AAXYAZ\to AXYAZ\to AXAYZ\to AAXYZ\to AXYZ$$

is equal to:

$$AXAYAZ\to AXAAYZ\to AXAYZ\to AAXYZ\to AXYZ$$

proceed:

$(AXAYAZ\to AAXYAZ\to AXYAZ)\to (AXAYZ\to AAXYZ\to AXYZ)=^{Ax}
=(AXAYAZ\to XAAYAZ\to XAYAZ\to AXYAZ) \to (AXAYZ\to XAAYZ\to XAYZ\to AXYZ)=^{natuality}$
$=AXAYAZ\to XAAYAZ\to (XAYAZ \to XAAYZ\to XAYZ\to AXYZ)=^{Ax}$
$=AXAYAZ\to XAAYAZ\to (XAYAZ \to XYAAZ\to XYAZ\to XAYZ\to AXYZ)=^{Ax}$
$=A_1XA_2YA_3Z\to XA_1A_2YA_3Z\to XA_{1,2}YA_3Z \to XA_{1,2}A_3YZ\to XAYZ\to AXYZ=^{naturality}$
$=A_1XA_2YA_3Z\to XA_1A_2YA_3Z\to XA_1A_2A_3YZ\to XA_{1,2}A_3YZ \to XAYZ\to AXYZ=^{monoid}$
$=A_1XA_2YA_3Z\to XA_1A_2YA_3Z \to XA_1A_2A_3YZ\to XA_1A_{2,3}YZ \to XAYZ\to AXYZ=^{naturality}$
$=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to XA_1A_2A_3YZ \to XA_1A_{2,3}YZ \to XAYZ\to AXYZ=^{naturality}$
$=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to (A_1XA_{2,3}YZ\to XA_1A_{2,3}YZ \to XAYZ\to AXYZ)=^{Ax}$
$=A_1XA_2YA_3Z\to A_1XA_2A_3YZ\to A_1XA_{2,3}YZ\to A_1A_{2,3}XYZ \to AXYZ.$

About the unity axioms we prove that $F_A(X)\cong IF_A(X)\to F_A(I)F_A(X)\to F(IX)\cong F(X)$ is the identity, this is::

$AX\cong IAX\to AAX\cong AIAX\cong IAAX\to IAX\cong AX$ for naturality this is:

$AX\cong IAX\to AAX\cong AIAX\cong IAAX\cong AAX\to AX$ for the usual monoidal topic this is :

$AX\cong IAX\to AAX\to AX=1$. The other unitary axiom is quite similar.