Question about proof of $A[X] \otimes_A A[Y] \cong A[X, Y] $ As far as I understand universal properties, one can prove $A[X] \otimes_A A[Y] \cong A[X, Y] $ where $A$ is a commutative unital ring in two ways:
(i) by showing that $A[X,Y]$ satisfies the universal property of $A[X] \otimes_A A[Y] $
(ii) by using the universal property of $A[X] \otimes_A A[Y] $ to obtain an isomorphism $\ell: A[X] \otimes_A A[Y] \to A[X,Y]$
Now surely these two must be interchangeable, meaning I can use either of the two to prove it. So I tried to do (i) as follows:
Define $b: A[X] \times A[Y] \to A[X,Y]$ as $(p(X), q(Y)) \mapsto p(X)q(Y)$. Then  $b$ is bilinear. Now let $N$ be any $R$-module and $b^\prime: A[X] \times A[Y] \to  N$ any bilinear map. 
I can't seem to define $\ell: A[X,Y] \to N$ suitably. The "usual" way to define it would've been $\ell: p(x,y) \mapsto b^\prime(1,p(x,y)) $ but that's not allowed in this case. 
Question: is it really not possible to prove the claim using (i) in this case?

Define $l(X^i Y^j) := b'(X^i, Y^j)$ and continue this to an $A$-linear map $A[X,Y] \to N$ by
$$l\left(\sum_{i,j} a_{ij} X^i Y^j\right) = \sum_{i,j} a_{ij} b'(X^i,Y^j).$$
