**EDIT**. A colleague wrote to me to point out that my original answer to this question was actually completely wrong: I had confused "isobaric" representations with "pure" representations (which are not at all the same thing).

The correct answer, according to my colleague, is this. Consider the one-dimensional representation $\sigma = |\cdot|^{1/2} \boxtimes|\cdot|^{-1/2}$ of $(GL_1 \times GL_1)(\mathbf{A})$. Then we can parabolically-induce this up to a representation $I(\sigma)$ of $GL_2(\mathbf{A})$. The representation $I(\sigma)$ is very highly reducible: its irreducible factors are precisely the representations of the form $\pi = \prod_v \pi_v$ where all but finitely many $\pi_v$ are the one-dimensional trivial representation of $GL_2(\mathbf{Q}_v)$, and the remainder are the Steinberg representation. All of these $\pi$'s are automorphic, but only the globally trivial representation (with no Steinberg places) is isobaric in Langlands' sense.

Notice that any two representations occurring in $I(\sigma)$ are isomorphic locally almost everywhere, so there is no "strong multiplicity one" for non-cuspidal automorphic reps. The point of isobaric representations seems to be to cut down from all automorphic reps to a smaller class in which there is some hope of multiplicity-one results being true.

*Original answer*: You can already see this for $GL_2$, so let me give an answer there, rather than for $GL_3$.

Consider the function $L(s) = \zeta(s - r +\tfrac{1}{2}) \zeta(s + r - \tfrac{1}{2})$, for an integer $r \ge 2$. This is the $L$-function of a non-cuspidal automorphic representation of $GL_2$ (given by the holomorphic Eisenstein series $E_{2r}$). However, it is not isobaric -- on the Galois side, it corresponds to a direct sum of two $\ell$-adic Galois representations which are each pure, but with different weights, while isobaric representations have all components pure of the same weight. ("Same weight" is the literal meaning of the word "isobaric".)