Let us assume that $1_{[-1,1]}$ is the identity function for $t$ but here I consider a generic function $f(t)$. Let us consider the Dirichlet problem on a bounded domain $D$

$$\Delta\phi_n+\lambda_n\phi_n=0 \qquad \phi=0\ on\ \partial D$$

You can write down the exact solution to your equation as

$$u(t,x,y)=\sum_n a_n(t)\phi_n(x)\phi_n(y)$$

so that $u(0,x,y)=\delta(x-y)$ implies $a_n(0)=1$. By a direct substitution you get the equations to be solved

$$\dot a_n+\lambda_na_n(t)+f(t)a_n(t)=0$$

that admits the solution

$$a_n(t)=e^{-\lambda_n t-\int_0^t dt'f(t')}.$$

In this way you should be able to get a better bound on the solution.

Now, let us assume that $1_{[-1,1]}$ is the identity function for $x$. The problem is reduced to the one of a Schroedinger equation for a rectangular potential barrier. Let us search for eigenfunctions to the problem

$$\partial^2\phi_E(x)-1_{[-1,1]}\phi_E(x)=-E\phi_E(x)$$

We expect a continuous spectrum in this case and will get

$$\phi^L_E(x)=A_1e^{ik_0x}+A_2e^{-ik_0x}\qquad x<-1$$
$$\phi^C_E(x)=B_1e^{ik_1x}+B_2e^{-ik_1x}\qquad x\in [-1,1]$$
$$\phi^R_E(x)=C_1e^{ik_0x}+C_2e^{-ik_0x}\qquad x>1$$

being $k_1=\sqrt{E-1}$ for $x\in [-1,1]$ and $k_0=\sqrt{E}$ otherwise. You now impose a continuity condition on the derivative and the eigenfunctions to get the coefficients. The final solution will take an integral form as

$$u(t,x,x_0)=\int_C dEe^{-Et}\phi_E(x)\phi_E(x_0)$$

with a properly chosen contour $C$. Please, note that is also $1_{[-1,1]}=\theta(x+1)-\theta(x-1)$ being $\theta(x)$ the Heaviside function.