Nate's suggestion on math.SE works. We'll show that if $A = k[x, \partial_x]$ and $B = k[y, \partial_y]$ are both taken to be the Weyl algebra, then the module over $A_2 = A \otimes B \cong k[x, \partial_x, y, \partial_y]$ generated by $e^{xy}$ is 1) simple and 2) not a tensor product of simple modules of $A$ and $B$.
Explicitly this module $M$ consists of elements of the form $f(x, y) e^{xy}$ where $f$ is a polynomial, with the obvious action by multiplication and differentiation. Abstractly it is the quotient of $k[x, \partial_x, y, \partial_y]$ by the left ideal $(x - \partial_y, y - \partial_x)$. We can show very straightforwardly that every nonzero element of $M$ generates it, by computing that
$$(\partial_x - y) x^i y^j e^{xy} = ix^{i-1} y e^{xy}$$
and similarly that
$$(\partial_y - x) x^i y^j e^{xy} = j x^i y^{j-1} e^{xy}.$$
In other words, this module is the pullback of the usual module $k[x, y]$ under the (edit: inverse of the) automorphism $A_2 \to A_2$ sending $x \mapsto x, y \mapsto y, \partial_x \mapsto \partial_x - y, \partial_y \mapsto \partial_y - x$. Now $k[x, y]$ is irreducible which means so is this module (explicitly, every nonzero element is a generator because we can repeatedly differentiate resp. apply the above maps to get to $1$ resp. $e^{xy}$), and $k[x, y]$ is a tensor product $k[x] \otimes k[y]$.
But we can show that $M$ is not such a tensor product (this property is not invariant under twisting by automorphisms). If it were such a tensor product $V \otimes W$, then a pure vector $v \otimes w$ would have the property that there is some differential operator $a(x, \partial_x) \in k[x, \partial_x]$ such that $(a \otimes 1)(v \otimes w) = av \otimes w = 0$; in words, it would satisfy a polynomial differential equation involving $x$ only.
No nonzero element of $M$ has this property. The key point is that every nonzero element of $k[x, \partial_x]$ makes elements bigger in the lex order: we have
$$x \left( x^i y^j e^{xy} \right) = x^{i+1} y^j e^{xy}$$
$$\partial_x \left( x^i y^j e^{xy} \right) = ix^{i-1} y^j e^{xy} + x^i y^{j+1} e^{xy}$$
and so formally, writing an arbitrary element $a \in k[x, \partial_x]$ as a sum $\sum a_{k, \ell} \partial_x^k x^{\ell}$, we see that if $\partial_x^k x^{\ell}$ is the largest monomial in $a$ in the lex order where $x > \partial_x$, then $a(x^i y^j e^{xy})$ has largest monomial $x^{i + \ell} y^{j + k}$, and in particular it does not vanish, so the same is true if $x^i y^j e^{xy}$ is replaced by any other element with the same largest monomial.
