Is the homomorphic image of an ideal an ideal?

No, it is not: let $i : \Bbb Z \to \Bbb Q$ be the natural injection. The only ideals of $\Bbb Q$ are $0$ and $\Bbb Q$. Take any ideal $(n)$ with $n \ne 0$ - is its image any of those two ideals of $\Bbb Q$? No, of course, so the image of an ideal is not necessarily ideal.

Since a ring homomorphism is a homomorphism of the underlying abelian groups under addition, $f(J)$ is an additive subgroup of $B$ (maps of abelian groups send subgroups to subgroups).

However, in general, the multiplicative absorption property of ideals need not be transferred: not every element of $B$ need be in the image of $f$, so showing that $f(x)f(y)\in f(J)$ for $x\in A$, $y\in J$ is not enough. You would need to show that for any $b\in\color{red}{B}$ and $x\in J$, $b f(x)\in f(J)$, and this is not generally true.

As others note, the image $f(J)$ is only an ideal of the subring $f(A)$ of $B$, not an ideal of $B$ (this is what you actually proved). It is not hard to find examples of morphisms of rings sending ideals to sets which are not ideals, as in Alex M.'s answer.

It is not true that the homomorphic image of an ideal is an ideal in general. Take any ring $R$ that can be embedded into a field $F$ (such as $\mathbb{Z}$ in $\mathbb{R}$). Then the image of a proper ideal $J \subset R$ under the embedding map will not be an ideal of $F$, as there are no proper nonzero ideals in a field.

In your proof, you never checked that if $a \in A$ and $b \in B$, then $bf(a) \in f(A)$, which is also necessary for an ideal.