For a nonunital commutative ring $A$, an $A$-module $M$ is called finitely generated over $A$ if there is a finite set of elements $x_1,...,x_n\in M$ such that every element of $M$ is of the form $a_1x_1+...+a_nx_n$ for some $a_1,...,a_n\in A$. Under this definition, notice that the ring $A$ may not be finitely generated over itself, for example $A=2\Bbb Z$.
Now let us consider a equivalent form of Nakayama's Lemma below:
For a nonunital commutative ring $A$ and a finitely generated module $M$ over $A$, there is an element $a\in A$ such that $am=m$ for all $M$.
The proof goes as usual: Write the generator $x_i=a_{i,1}x_1+...+a_{i,n}x_n$ and notice the determinant of the matrix $\phi=\det(a_{i,j}-\delta_{i,j}\Bbb{1}_M)$ vanishes on every element of $M$ while $\phi$ is of the form $\Bbb{1}_M+k\Bbb{1}_M$ for some $k\in A$(just expand out the $\det$.)
To obtain the usual version of Nakayama's lemma, simply replace $A$ by an ideal $\mathfrak a$ in a commutative unital ring with $\mathfrak aM=M$.
I am requesting another proof of this fact using unitalization, as I was told the category of all $A$-modules is indeed isomorphic to the category of all $U(A)$-modules, where $U(A)$ is the unitalization of $A$. I really cannot come up with another categorical proof of the nonunital version of Nakayama's Lemma, although this "nonunital" version involves less restriction while seems indeed fundamental in its form. I will be appreciated if another angle is possible to attack this result.
You can just reverse this argument. $A$ is an ideal in its unitization $U(A)$, and considering $M$ as a $U(A)$-module, it satisfies $AM=M$. So, the usual version of Nakayama's lemma gives an element $a\in A$ such that $am=m$ for all $m\in M$.