$\newcommand\Th\Theta\newcommand{\Z}{\mathbb Z}$The answer here is no. 

E.g., for $(x,t)\in[0,1]\times[0,\infty)$ let 
\begin{equation}
	f(x,t):=2x\,1(x<1/2)+(2-2x)\,1(x\ge1/2). 
\end{equation}
Then 
\begin{equation}
	(\Th*f)(x,t)=\frac t4-\frac2{\pi^3}\sum_{n=1}^\infty a_n\cos2n\pi x
	=\frac t4-\frac{c_t}{\pi^3}\,g_t(x), 
\end{equation}
where 
\begin{equation}
	a_n:=a_{t,n}:=(1+(-1)^{n-1})\frac{1-e^{-\pi n^2t/2}}{n^4}, 
\end{equation}
\begin{equation}
	g_t(x):=\sum_{n\in\Z}p_n e^{2\pi ixn},  
\end{equation}
$p_n:=a_n/c_t$ for $n\ne0$, $p_0:=0$, 
$c_t:=2\sum_{n=1}^\infty a_n$, so that $p_n\ge0$ for all $n$ and $\sum_{n\in\Z} p_n=1$. 

So, $g_t$ is the characteristic function (c.f.) of a random variable $X_t$ such that $P(X_t=2\pi n)=p_n$ for $n\in\Z$. Note that $EX^4=\infty$. So, the fourth derivative of $g_t$ at $0$ does not exist (see e.g. [Theorem 2.3.1][1]). So, $(\Th*f)(x,t)$ is not smooth in $x$ at $x=0$ and hence is not jointly smooth in $(x,t)\in[0,1]\times(0,\infty)$. $\quad\Box$ 

[1]: https://www.amazon.com/Characteristic-functions-Eugene-Lukacs/dp/0852641702