$\newcommand\si{\sigma}\newcommand\la{\lambda}\newcommand\th{\theta}\newcommand\Th{\Theta}$Suppose that your family $(f_\th)$ is an exponential family, where $\th:=(\lambda,\sigma)\in\Th:=(0,\infty)^2$. Since $f_\th>0$ for all $\th\in\Th$, we have $h>0$. So, we can define $$g(\th,x):=\ln f_\th(x)=\eta(\th)T(x)-A(\th)+\ln h(x)$$ for real $x$.
Take any $\th_0\in\Th$ and let $$G_{\th_0}(\th,x):=g(\th,x)+g(\th_0,0)-g(\th,0)-g(\th_0,x) =\xi(\th)S(x),$$ where $\xi(\th):=\eta(\th)-\eta(\th_0)$ and $S(x):=T(x)-T(0)$. So, for any $\th_1$ and $\th_2$ in $\Th$ and all real $x_1$ and $x_2$ we will have $$G_{\th_0}(\th_1,x_2)G_{\th_0}(\th_2,x_1)=G_{\th_0}(\th_1,x_1)G_{\th_0}(\th_2,x_2). \tag{1}\label{1}$$
To show that your family $(f_\th)$ is not an exponential family, it remains to find $\th_0,\th_1,\th_2$ in $\Th$ and real $x_1$ and $x_2$ such that \eqref{1} fails to hold. This should be very easy to do, say by taking $\th_0,\th_1,\th_2$ in $\Th$ and real $x_1$ and $x_2$ "at random". For instance, if $\th_0=(1,1)=\th_1$, $\th_2=(2,1)$, $x_1=1$, and $x_2=2$, then the left- and right-hand sides of \eqref{1} are $-0.102\ldots$ and $0.075\ldots$, respectively, so that \eqref{1} fails to hold.