I think the answers in the comments are slightly besides the point. It's true that the integrals aren't well-defined in the case of $g(t)=t$, but the more fundamental point here seems to me that
$\int_{-\infty}^{+\infty} g(t-a)\mathrm{d}a=\int_{-\infty}^{+\infty} g(a)\mathrm{d}a$
is in fact true, and whenever one of the sides is well-defined the other is, too. It follows without any mention of convolutions from a simple change of variables $u=t-a$, which leaves the limits of integration at infinity unchanged. So it seems that to answer the question needs to involve addressing why Joey thought that these two were not equal in the first place, which probably had nothing to do with lack of convergence. Joey, I'm guessing that you didn't see that making the substitution from $a$ to $t-a$ leaves the limits at infinity unchanged?
Another way of seeing this is that the area under $g(t-a)$ is just the shifted mirror image of the area under $g(a)$.