I've been working on this problem and was able to get it down to this inequality:
$$E[\log X_t] \leq ae^{-t} + b(1-e^{-t}) + e^{-t}\int_0^t e^sE[\log X_s] ds$$
I'd like to use Gronwall's Inequality to determine an expression for $\limsup E[\log X_t]$
So using Gronwall's Inequality, I get:
$$E[\log X_t] \leq ae^{-t} + b(1-e^{-t}) + \int_0^t \left[ae^{-s} + b(1-e^{-s})\right]e^se^{\int_s^t e^r dr}$$
So the $``ae^{-t} + b(1-e^{-t})"$ -term outside the integral is easy. Inside the integral is where I seem to get stuck.
The integral becomes:
$$\int_0^t \left[a + b(e^s-1)\right]e^{e^t-e^s} ds = e^{e^t}\int_0^t \frac{a-b}{e^{e^s}} + \frac{be^s}{e^{e^s}} ds$$
I have no idea how to compute that. I don't really need an explicit answer but I need to compute the $\limsup_{t \rightarrow \infty}$ of that expression. I don't see a way to do this.
This just a question about the real valued function $u(t) = E[\log X_t]$. Your inequality says $$ u(t) \le ae^{-t} + b(1-e^{-t}) + e^{-t}\int_0^t e^s u(s) ds $$ or equivalently with $v(t) = e^tu(t)$ $$ v(t) \le a + b(e^t-1) + \int_0^t v(s) ds $$ Now Gronwall's Lemma implies that $$ v(t) \le a + b(e^t - 1) + \int_0^t (a + b(e^s-1))e^{t-s}ds \\ = e^t(a+b(t-1)) +b-a \, . $$ Therefore, $$ E[\log X_t] = u(t) \le e^{-t} v(t) = (a+b(t-1)) + (b-a)e^{-t} \, . $$ Unfortunately, this does not give a $t$-independent bound for $\limsup_t E[\log X_t]$.