Convolution must be a bounded bilinear operator if it is well-defined

Under the assumption of your question, the bilinear operator $T: L^p \times L^1 \to L^1$ by $T(f,g) = f \ast g$ is well-defined. Additionally, define $T^f:L^1 \to L^1$ and $T_g:L^p \to L^1$ for $f \in L^p$ and $g \in L^1$ by setting $T^f(g) = T(f,g) = T_g(f)$. I proceed in several steps.

Step 1: $T_g$ is bounded

This is very similar to the classic application of the UBT that you mention. Mimicking that application, set $$g_n(x):=\begin{cases} n, & \lvert g(x) \rvert \ge n\ \text{and } |x|<n, \\ g(x), & \lvert g(x)\rvert <n\ \text{and }|x| < n,\\ 0,& |x| \geq n. \end{cases}$$

By the closed graph theorem, each $T_{g_n}$ is a bounded operator. Indeed, suppose that $f_k \to f$ in $L^p$ and $T_{g_n} f_k \to h$ in $L^1$. Then notice that \begin{align*} \left | \int (f(y) - f_k(y)) g_n(x-y) dy \right | \leq \|f-f_k\|_{L^p} \|g_n\|_{L^{p'}} \leq C_n \|f-f_k\|_{L^p} \to 0 \end{align*} as $k \to \infty$. This means that $T_{g_n} f_k \to T_{g_n} f$ pointwise as $k \to \infty$ and so $h = T_{g_n} f$.

Also, we have that $|T_{g_n}f| \leq T(|f|,|g|)$ pointwise and $T(|f|,|g|) \in L^1$ by assumption. Therefore, by an application of the uniform boundedness theorem, $C_1 := \sup_n \|T_{g_n}\| < \infty$.

To conclude this step, it remains to see that $T_{g_n}f \to T_g f$ in $L^1$ as $n \to \infty$. For this, first notice that $$|f(x-\cdot) [g_n(\cdot) - g(\cdot)] | \leq 2 |f(x-\cdot) g(\cdot)|$$ and since $T(|f|,|g|) < \infty$ a.e. the right hand side is in $L^1$ for almost all $x$. Hence we can apply the dominated convergence theorem to see that $T_{g_n}f \to T_gf$ a.e. Then using the fact that $|Tg_nf - T_g f| \leq 2 T(|f|,|g|)$ we can apply the dominated convergence theorem again to see that $T_{g_n} f \to T_g f$ in $L^1$.

Step 2: $T^f$ is bounded

This is basically the same argument as above. Define $$f_n(x):=\begin{cases} n, & \lvert f(x) \rvert \ge n\ \text{and } |x|<n, \\ g(x), & \lvert f(x)\rvert <n\ \text{and }|x| < n,\\ 0,& |x| \geq n. \end{cases}$$ The argument then runs almost line for line the same as in step $1$ with the roles of $f$ and $g$ reversed, except that in the application of the closed graph theorem you now have $p = 1$ and $p' = \infty$ (which causes no issues at all).

Step 3: The conclusion

This is now a standard application of the UBT. Consider the set $U = \{T_g : \|g\|_{L^1} = 1\}$. Then for each $g$ with $\|g\|_{L^1} = 1$, $$\|T_g f\| = \|T^f g \| \leq \|T^f\|$$ so that by the UBT, $C_2 = \sup_{\|g\|_{L^1} = 1} \|T_g\| < \infty$. Hence for arbitrary $f \in L^p$ and $g \in L^1$ $$\|T(f,g)\|_{L^1} = \|g\|_{L^1} \|T_{\frac{g}{\|g\|_{L^1}}} f \| \leq C_2 \|g\|_{L^1} \|f\|_{L^p}$$ as desired.

First some abstract stuff: Suppose $E,F,G$ are locally convex spaces such that $F\subset G$ (with continuous embedding) and $T:E\rightarrow G$ is a continuous linear map with $T(E)\subset F$. If $E$ and $F$ are Fréchet, then the closed graph theorem implies that $T$ is automatically continuous as map $T:E\rightarrow F.$ A similar argument works for a bilinear map $B:E_1 \times E_2 \rightarrow G$ with $B(E_1,E_2)\subset F$, applying the linear result to $B(x,\cdot)$ and $B(\cdot, y)$ and noting that if $E_1,E_2$ and $F$ are Fréchet, then separate continuity in each variable implies joint continuity.

Hence, if you can show that convolution is continuous as map $L^1 \times L^p\rightarrow G$ for some locally convex space $G\supset L^1$, then the assumption $L^1\ast L^p \subset L^1$ and the abstract nonsense from above already imply continuity into $L^1$. I suppose that $G= \mathcal{D}'(\mathbb{R})$ should work but I have not worked that out.