The case $x=0$ can be deduced from the identity in law due to Paul Lévy between $(S_t-W_t,S_t)$ and $(|W_t|,L(t,0))$, where $S_t$ is the maximum of $W$ on $[0,t)$. Hence $P(L(t,0)\ge z)=P(S_t\ge z)$ and, since $P(S_t\ge z)=2P(W_t\ge z)$ by Désiré André's reflexion principle, the density of the distribution of $L(0,x)$ is twice the density of $W_t$ restricted to $(0,+\infty)$. That is, writing $g(t,\cdot)$ for the centered Gaussian density with variance $t$, $$ P(L(t,0)\in\mathrm{d}u)=2g(t,u)\mathbf{1}_{u > 0}\mathrm{d}u. $$ If $x\ne0$, $L(t,x)$ is distributed like $L((t-\tau_x)^+,0)$ where $\tau_x$ is independent of $W$ and distributed like the first hitting time of $x$ by $W$ (this is the strong MrkovMarkov property of $W$ at time $\tau_x$). The distributions of $\tau_x$ and $\tau_{-x}$ coincide and the distribution of $\tau_x$ is known since $[\tau_x\le t]=[S_t\ge x]$ for every $x>0$. This yields, in principle, the distribution of $L(t,x)$ as a barycenter of distributions of random variables $L(s,0)$ for $s$ in $[0,t)$ but I doubt one can go much further than writing the density of $L(t,x)$ more nicely than as an integral.
Unless I am mistaken, for every $x\ne0$, one gets $$ P(L(t,x)\in\mathrm{d}u)=p_{t,x}\delta_0(\mathrm{d}u)+h_{t,x}(u)\mathbf{1}_{u > 0}\mathrm{d}u, $$
where $$p_{t,x}=P(L(t,x)=0)=P(S_t < |x|)=1-2P(W_t > |x|)=2\int_0^{|x|}g(t,y)\mathrm{d}y. $$ and, writing $\partial_s$ for the partial derivative with respect to $s$, for every positive $u$, $$h_{t,x}(u)=4\int_0^tg(t-s,u)\mathrm{d}s\int_{|x|}^{+\infty}\partial_sg(s,y)\mathrm{d}y, $$
In fact, rather than considering a fixed time $t$, one often looks at the distribution of $L(Z,x)$, where $Z$ is a random time independent on $W$ and exponentially distributed. In other words, one considers the Laplace transform (with respect to the time argument) of the random function $L(\cdot,x)$.
All this is done in many places, see for example thea set of lecture notes on Brownian motion by Peter Mörters and Yuval Peres, available here (chapter 2).