$\newcommand\Th\Theta\newcommand{\Z}{\mathbb Z}$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). So, $(\Th*f)(x,t)$ is not smooth in $x$ at $x=0$ and hence is not jointly smooth in $(x,t)\in[0,1]\times(0,\infty)$. $\quad\Box$