Suppose we have an algebraic morphism $f:\mathbb{A}^2\rightarrow \mathbb{A}^2$. Can the image of $f$ be the zero locus of the polynomial $xy$?
I think not, at least not if we're working over $\mathbb{C}$ (or any algebraically closed, hence infinite, field). My reasoning is that $f$ could be written as $f=(f_1,f_2)$ with $f_1,f_2\in\mathbb{C}[s,t]$. If $Im(f)=V(xy)$, $f_1f_2$ would have infinitely many zeroes, i.e. it would be the 0 polynomial, and therefore either $f_1$ or $f_2$ would be 0. But the points $(0,1)$ and $(1,0)$ cannot both lie in the image of $f$, which gives a contradiction.
Can somebody please tell me if this reasoning is correct? And is there a nice counterexample in the case that the field over which we work is not algebraically closed?
In your argument, it's not clear to me why $f_1f_2(s,t)$ must have finitely many zeros. Zero sets of nontrivial single variable polynomials are finite, but those of multivariate polynomials need not be.
However, it's still true that this is not a possible image for $\mathbb A^2$, because an irreducible algebraic set cannot surject onto a reducible one. Intuitively, this is similar to the fact that the image of a connected topological space is connected. With this in mind, let's solve your problem by proving the more general statement (assuming $k$ to be algebraically closed throughout).
Let $X=Z(I)$ for $I\subset k[x_1,\dots,x_m]$ and $Y=Z(J)$ for $J \subset k[y_1,\dots,y_m].$ Assume that $I$ and $J$ are radical. Recall that algebraic morphisms $f:X \rightarrow Y$ are in one-to-one correspondence with $k$-algebra homomorphisms $f^{\#}:k[y_1,\dots,y_m]/J \rightarrow k[x_1,\dotsm,x_m]/I$. Recall also that if $f$ is surjective then $\ker f^{\#}=\{0\}$.
Now suppose that $X$ surjects onto $Y$, and that
The above discussion implies that $k[y_1,\dots, y_m]/J$ embeds into $k[x_1,\dots,x_n]/I$. Since $J$ is not prime, $k[y_1,\dots,y_m]/J$ is not a domain. Since $I$ is prime, $k[x_1,\dots,x_n]/I$ is a domain. Hence, this is a contradiction.
For your problem, simply check that $0=I \subset k[s,t]$, and $(xy)=J \subset k[x,y]$ satisfy $(*)$, so there is no surjection $\mathbb A^2 \rightarrow Z(xy)$.
Edit: here's a counterexample for $k$ not algebraically closed. Take $k=\mathbb Z_2$, and consider $f:k^2 \rightarrow k^2$ given by $(x,y) \mapsto (xy,x+1)$. Then $Z(xy)=\{(0,0),(0,1),(1,0)\}$, and the image of $f$ is precisely $Z(xy)$:
$$(0,0) \mapsto (0,1),$$
$$(0,1) \mapsto(0,1),$$ $$(1,0) \mapsto (0,0),$$ $$(1,1) \mapsto (1,0).$$