No. Let's take $f(x)=e^{-x}$, $$ g(x) = \int e^{-tx}\, d\mu(t) $$ with $\mu$ supported by $(0,2)$. We must take $\mu$ such that $\mu(0,2)=1$, $\int t\, d\mu(t)<1$, and $\mu(0,1)>0$. Then we will have the desired inequality $f\le g$ near zero and infinity.

Now let's try to make $f(\delta)=e^{-\delta}> g(\delta) = \int e^{-\delta t}\, d\mu(t)$ for some small $\delta>0$. I'm giving up.

No. Let's take $g(x)=\epsilon e^{-\epsilon x}+(1-\epsilon)e^{-\alpha x}$, $$ f(x) = \int_{\epsilon}^{1+\epsilon} e^{-tx}\, dt = \frac{1}{x}e^{-\epsilon x}(1-e^{-x}) . $$ Then clearly $f\le g$ near infinity, and near zero, $f(x)\simeq 1-(1/2+\epsilon)x$, $g(x)\simeq 1-(\epsilon^2+(1-\epsilon)\alpha)x$. I also want $f\le g$ near zero, and this gives a condition on $\alpha$ (given $\epsilon$). For small $\epsilon$, I can take $\alpha\approx 1/2$.

On the other hand, as $\epsilon\to 0$, $\alpha\to 1/2$, we have that $f(1)\to 1-e^{-1}$, $g(1)\to e^{-1/2}$, and now a calculator will tell us that $f(1)>g(1)$ for sufficiently small $\epsilon$. (There might be easier counterexamples.)

No. Let's take $f(x)=e^{-x}$, $$ g(x) = \frac{1}{L} \int_0^L e^{-tx}\, dt = \frac{1}{Lx}(1-e^{-Lx}) , $$ with $L\gg 1$. Then clearly $f\le g$ near infinity, and $f(1/L)=e^{-1/L}>g(1/L)=1-e^{-1}$. Near $x=0$, we observe that $f'(0)=g'(0)=-1$, so a small modification of $g$ will make sure that $f\le g$ near zero also. More specifically, we take $g=\int e^{-tx}\, d\mu(t)$, where $\mu$ is almost uniform on $(0,L)$, but a tiny amount of mass has been moved from near zero to near $L$. (There might be easier counterexamples.)

No. Let's take $f(x)=e^{-x}$, $$ g(x) = \int e^{-tx}\, d\mu(t) $$ with $\mu$ supported by $(0,2)$. We must take $\mu$ such that $\mu(0,2)=1$, $\int t\, d\mu(t)<1$, and $\mu(0,1)>0$. Then we will have the desired inequality $f\le g$ near zero and infinity.

Now let's try to make $f(\delta)=e^{-\delta}> g(\delta) = \int e^{-\delta t}\, d\mu(t)$ for some small $\delta>0$. I'm giving up.