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?
 A: New version of the answer
I'll leave below the old version of the answer so that the comments remain understandable. Thank you to the commenters. 
Let me prove a slightly more general statement:
(I) If $(X_i)_{i\in I}$ is a family of indeterminates, then we have a natural $A$-algebra isomorphism 
$$
A\left[(X_i)_{i\in I}\right]\simeq\bigotimes_{i\in I}A[X_i].
$$ 
Here and in the sequel, tensor products are taken over $A$.
Let $B$ and $C$ be the left and right hand side of the above display, and recall that the coproduct of a family $(A_i)_{i\in I}$ of $A$-algebras is their tensor product. Also note:
(II) For any $A$-algebra $D$ and any family $(d_i)_{i\in I}$ a family of elements of $D$, there is a unique $A$-algebra morphism from $B$ to $D$ mapping $X_i$ to $d_i$ for all $i$.
(III) For any $A$-algebra $D$ and any family $(d_i)_{i\in I}$ a family of elements of $D$, there is a unique $A$-algebra morphism from $C$ to $D$ mapping 
$$
x_i:=X_i\otimes\bigotimes_{j\neq i}1
$$
to $d_i$ for all $i$.
Proof of (I): By (II) and (III) the $A$-algebra morphism from $B$ to $C$ mapping $X_i$ to $x_i$ for all $i$ and the $A$-algebra morphism from $C$ to $B$ mapping $x_i$ to $X_i$ for all $i$ are inverse isomorphisms.
Additional remarks. 
$\bullet$ Inductive limits of $A$-algebras exist. In the terminology of 
Categories and Sheaves by Kashiwara and Schapira (Springer 2006), Google preview, Amazon preview,
we have: Let $\mathcal U$ be a universe, let $\mathcal A$ be the category of $A$-algebras whose underlying set belongs to $\mathcal U$, let $I$ be a small category, and let $\alpha:I\to\mathcal A$ be a functor. Then the inductive limit of $\alpha$ exists in $\mathcal A$.
$\bullet$ If $(M_i)_{i\in I}$ is a family of $A$-modules, the the tensor product $\bigotimes_{i\in I}M_i$ is well defined, and satisfies the usual universal property.
Old version of the answer 
$A[X,Y]$ and $A[X]\otimes_AA[Y]$ can viewed be 
as rings, as $A$-algebras, as $A$-modules, as $A[X]$-algebras, as $A[X]$-modules, as $A[Y]$-algebras, as $A[Y]$-modules, and in many others ways. I'll view them as $A$-algebras. Then the natural isomorphism $A[X,Y]\simeq A[X]\otimes_AA[Y]$ is an immediate consequence of the following two more general facts.
If $B$ and $C$ are $A$-algebras, let me denote by $\mathcal A(B,C)$ the set of $A$-algebra morphisms from $B$ to $C$.
Fact $1$. If $B,C,D$ are $A$-algebras, then $B\otimes_AC$ is the coproduct of $B$ and $C$, that is, we have a canonical bijection 
$$
\mathcal A(B\otimes_AC,D)\simeq\mathcal A(B,D)\times\mathcal A(C,D).
$$
Fact $2$. If $(X_i)_{i\in I}$ is a family of indeterminates, then $A\left[(X_i)_{i\in I}\right]$ is free over $I$, that is we have a canonical bijection
$$
\mathcal A\left(A\left[(X_i)_{i\in I}\right],D\right)\simeq D^I,
$$
where $D^I$ is the set of maps from $I$ to $D$.
EDIT A. This is to answer Marc's comment. Let $f:B\to C$ be an $A$-algebra morphism. For any $A$-algebra $D$ let 
$$
f_D:\mathcal A(C,D)\to\mathcal A(B,D)
$$
be the induced map. Assume that $(1)$ $f_B$ is surjective and $(2)$ $f_C$ injective. Then, by $(1)$, there is a $g:C\to B$ such that $g\circ f=\text{id}_B$, and $(2)$ implies $f\circ g=\text{id}_C$. (This is a general trick.)
EDIT B. To complete the argument, one needs a morphism between $A[X,Y]$ and $A[X]\otimes_AA[Y]$. This is obtained by using the (omitted but obvious) description of the above canonical bijections.
A: 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
$$l\left(\sum_{i,j} a_{ij} X^i Y^j\right) = \sum_{i,j} a_{ij} b'(X^i,Y^j).$$
A: Why Not show that the tensor product has the universal property of the polynomial algebra? For me that is the more intuitive way of proving this fact.
A: For a commutative unital ring $R$ we can show the following $R$-module isomorphism in two ways: $R[x,y] \cong R[x] \otimes R[y]$.

(i) Obtain an isomorphism by using the universal property of $(R[x] \otimes R[y], b)$ where $b: R[x] \times R[y] \to R[x] \otimes R[y]$ is the bilinear map $(p(x), q(y)) \mapsto p(x) \otimes q(y)$ (it looks like this by construction of the tensor product). For $R[x,y]$ we observe that $b^\prime: R[x]\times R[y] \to R[x,y]$, $(p(x),q(y)) \mapsto p(x)q(y)$ is bilinear hence there exists a unique linear map $l: R[x] \otimes R[y] \to R[x,y]$ such that $l \circ b = b^\prime$. We claim that $l$ is an isomorphism, that is, has a two sided inverse. To see this, define $f: R[x,y] \to R[x] \otimes R[y]$ as $\sum a_{ij} x^i y^j \mapsto \sum a_{ij} (x^i \otimes y^j)$. Then 
$$ l (f (\sum a_{ij} x^i y^j)) = \sum a_{ij} x^i y^j$$ and 
$$ f (l (\sum a_{ij} (x^i \otimes y^j))) = \sum a_{ij} (x^i \otimes y^j)$$

(ii) Show that $R[x,y]$ together with the map $b: R[x] \times R[y] \to R[x,y]$, $(p(x) , q(y)) \mapsto p(x) q(y)$ satisfies the universal property of $R[x] \otimes R[y]$. That is, for an $R$-module $M$ and a a bilinear map $b^\prime : R[x] \times R[y] \to M$ there is a unique linear map $l: R[x,y] \to M$ such that $l \circ b = b^\prime$. Define $l: \sum a_{ij} x^i y^j \mapsto \sum a_{ij}b^\prime (x^i, y^j)$. Then $l$ is linear, $l \circ b = b^\prime$ and $l$ is unique: let $l^\prime$ be such that $l^\prime \circ b = b^\prime$. Then $l^\prime \circ b (x^i, y^j) = b^\prime(x^i, y^j) = l \circ b (x^i, y^j)$ and hence $l(x^i y^j) = l^\prime(x^i y^j)$ and hence $l = l^\prime$.
A: I can see your problem. As Marlu suggested and in some of the comments above, the trick is to treat of the elements $x^iy^j$ as a "basis" for the polynomial ring $A[x,y]$. In fact, this is the trick because suppose you take some $\sum_{i,j} X^iY^j \in A[x,y]$. Suppose hypothetically that you already have a linear map $L : A[X,Y] \rightarrow N$. Then the action of $L$ on this polynomial being $\sum_{i,j}L(X^iY^j)$ so that the image of any polynomial in $A[X,Y]$ is in fact completely determined by the action of $L$ on the $X^iY^j$. Let us keep this idea in mind and consider the diagram below.

Because we want the diagram to commute, we should have just concentrating on $X^iY^j$ that 
$$\begin{eqnarray*} b'(X^i,Y^j) &=& \ell \circ b(X^i,Y^j) \\
&=&\ell(X^iY^j) \end{eqnarray*} $$
Now from what I said in the first paragraph, you can extend $\ell$ additively. Let us check that $\ell$ is compatible with scalar multiplication. Take any $a \in A$. Then $$\begin{eqnarray*} \ell(aX^iY^j) &=& b'(aX^i,Y^j)\\
 &=& ab'(X^i,Y^J) \\ &=& a\ell(X^iY^j) \end{eqnarray*}$$
I could take the $a$ out of $b'(\cdot, \cdot)$ because we are now considering $A[X]$ and $A[Y]$ as $A$ - modules and so $b'$ is $A$ - bilinear. We have now completed the check that $\ell$ is linear and uniqueness should be obvious. It follows you have shown that $A[X,Y]$ satisfies the universal property of the tensor product $A[X] \otimes_A A[Y]$ from which it follows that 
$$A[X,Y] \cong A[X] \otimes_A A[Y].$$
$$\hspace{6in} \square$$
