Interesting question: obviously a sufficient condition for your operator $T$ to be a bounded endomorphism of $L^\infty$ is that $$ \sup_x\int\vert K(x,y)\vert dy=C<+\infty \ (\sharp).\quad\text{This implies trivially $\vert (Tu)(x)\vert\le C\Vert{u}\Vert_{L^\infty}$.} $$ Note that this is precisely the case for the convolution by an $L^1$ function $f$, a case for which $K(x,y)=f(x-y)$. However the most interesting part of the question is what you did not ask: Note that $$ \int_\mathbb R\vert K(x,y)\vert dy=\sup_{\Vert u\Vert_{L^\infty}=1}\Bigl\vert\int K(x,y) u(y) dy\Bigr\vert=\sup_{\Vert u\Vert_{L^\infty}=1}\vert (Tu)(x)\vert, $$ so that if $T$ is a bounded endomorphism of $L^\infty(\mathbb R)$, this implies $$ ess\sup_{x\in \mathbb R}\int_\mathbb R\vert K(x,y)\vert dy\le \Vert T\Vert<+\infty, $$ so that $(\sharp)$ is indeed an iff condition for $T$ to be a bounded endomorphism of $L^\infty(\mathbb R)$. Last but not least, many interesting operators fail to be bounded endomorphisms of $L^1$ and of $L^\infty$ but are bounded on $L^p$ for $p\in(1,+\infty)$. This is the case in particular of the Hilbert transform (convolution with $pv(1/x)$) which sends $L^1$ into $L^1_{weak}$ and $L^\infty$ into $BMO$ by a Marcinkiewicz argument.

Interesting question: obviously a sufficient condition for your operator $T$ to be a bounded endomorphism of $L^\infty$ is that $$ \sup_x\int\vert K(x,y)\vert dy=C<+\infty \ (\sharp).\quad\text{This implies trivially $\vert (Tu)(x)\vert\le C\Vert{u}\Vert_{L^\infty}$.} $$ Note that this is precisely the case for the convolution by an $L^1$ function $f$, a case for which $K(x,y)=f(x-y)$. Many interesting operators fail to be bounded endomorphisms of $L^1$ and of $L^\infty$ but are bounded on $L^p$ for $p\in(1,+\infty)$. This is the case in particular of the Hilbert transform (convolution with $pv(1/x)$) which sends $L^1$ into $L^1_{weak}$ and $L^\infty$ into $BMO$ by a Marcinkiewicz argument.