You have shown that $e^{-Ht}\colon L^1\to L^2$ is bounded with the given operator norm. It follows that the dual operator $$\left(e^{-Ht}\right)^*\colon (L^2)^* \to (L^1)^*$$ is bounded with the same operator norm. Since $(L^1)^* = L^\infty$ and $(L^2)^* = L^2$ (these are isomorphisms, actually), it remains to check that $\left(e^{-Ht}\right)^* = e^{-Ht}$.

For this note that $$\int f(x)(e^{-Ht}g)(x)dx = \iint f(x)e(x,y;t)g(y)dxdy$$ where $e(x,y;t)$ is the heat kernel, which is real and symmetric in the variables $x,y$. It follows that $\left(e^{-Ht}\right)^*$ is again integration against the heat kernel as claimed.

You have shown that $e^{-Ht}\colon L^1\to L^2$ is bounded with the given operator norm. It follows that the dual operator $$\left(e^{-Ht}\right)^*\colon (L^2)^* \to (L^1)^*$$ is bounded with the same operator norm, because of the characterisation $$\|T\|_{L^1\rightarrow L^2}=\|T^*\|_{L^\infty\rightarrow L^2}=\sup_{f,g}\frac{\langle Tf,g\rangle}{\|f\|_1\|g\|_2}\,.$$
Since $(L^1)^* = L^\infty$ and $(L^2)^* = L^2$ (these are isomorphisms, actually), it remains to check that $\left(e^{-Ht}\right)^* = e^{-Ht}$.

For this note that $$\int f(x)(e^{-Ht}g)(x)dx = \iint f(x)e(x,y;t)g(y)dxdy$$ where $e(x,y;t)$ is the heat kernel, which is real and symmetric in the variables $x,y$. It follows that $\left(e^{-Ht}\right)^*$ is again integration against the heat kernel as claimed.

You have shown that $e^{-Ht}\colon L^2\to L^1$ is bounded with the given operator norm. It follows that the dual operator $$\left(e^{-Ht}\right)^*\colon (L^1)^* \to (L^2)^*$$ is bounded with the same operator norm. Since $(L^1)^* = L^\infty$ and $(L^2)^* = L^2$ (these are isomorphisms, actually), it remains to check that $\left(e^{-Ht}\right)^* = e^{-Ht}$.

For this note that $$\int f(x)(e^{-Ht}g)(x)dx = \iint f(x)e(x,y;t)g(y)dxdy$$ where $e(x,y;t)$ is the heat kernel, which is real and symmetric in the variables $x,y$. It follows that $\left(e^{-Ht}\right)^*$ is again integration against the heat kernel as claimed.

You have shown that $e^{-Ht}\colon L^1\to L^2$ is bounded with the given operator norm. It follows that the dual operator $$\left(e^{-Ht}\right)^*\colon (L^2)^* \to (L^1)^*$$ is bounded with the same operator norm. Since $(L^1)^* = L^\infty$ and $(L^2)^* = L^2$ (these are isomorphisms, actually), it remains to check that $\left(e^{-Ht}\right)^* = e^{-Ht}$.

For this note that $$\int f(x)(e^{-Ht}g)(x)dx = \iint f(x)e(x,y;t)g(y)dxdy$$ where $e(x,y;t)$ is the heat kernel, which is real and symmetric in the variables $x,y$. It follows that $\left(e^{-Ht}\right)^*$ is again integration against the heat kernel as claimed.