Your questions shows some confusions about how orbits work. Kepler's three laws are useful in describing planetary orbits, but they don't quite get you the entire way into a direct way to calculate the position $\mathbf r(t)$ given a specified time $t$.
Instead, you require a solution of Newton's equations of motion for the planet. There are two ways to do this:
One is numerically: take a starting position (like a fixed starting ephemeris) and then propagate it numerically using the numerical ODE solver of your choice. This could be based on Euler's method, some level of the Runge-Kutta method, Verlet's algorithm as described in Rutger's answer. In general, it takes slightly more work to implement a higher-order method such as Runge-Kutta, but once you do it can result in surprisingly accurate results even for moderate time steps.
Alternatively, you can solve it analytically - or, of course, rely on the analytical solution that's been known since the time of Newton. The problem here is that the analytical method will only give you a closed-form, elementary expression for the distance $r(t)$ from the focus in terms of the angular variable $\theta(t)$, defined such that $\theta(t)=0$ at periapsis, in the form
$$r(t)=\frac{p}{1+e \cos(\theta(t))}, \tag1$$
where $p$ describes the size of the orbit and $e$ is the eccentricity. If you start with $t$ and you want the position, though, there is no closed-form elementary expression that you can use; instead, you're forced to solve a trascendental equation known as Kepler's equation, which reads
$$t=\kappa (E-e\sin(E)), \tag{2}$$
where $\kappa$ is a constant and $E$ is the eccentric anomaly, which gives you the angular coordinate via
$$\cos(\theta)=\frac{\cos(E)-e}{1-e\cos(E)}.$$
The problem is that Kepler's equation $(2)$ is a transcendental equation, which means that it doesn't have closed-form solutions and you will need to do it numerically, using e.g. Newton's method. This isn't terrible but it does mean that even the analytical solution has a nonzero numerical weight to it, and if you want to be doing this repeatedly and accurately you will end up spending a lot of time numerically solving $(2)$.
For more details, see the Wikipedia articles on Kepler orbits and the position as a function of time under Kepler's laws of motion.
If you just want one position at a single time, it is probably cheaper to implement the exact solution, which will have a nonzero but fairly trivial numerical cost. It is unlikely, however, that you do in fact want a single position, and more likely you want to plot a stretch of the orbit for a variety of times, in which case the numerical cost of the exact solution could be higher than a properly-implemented Runge-Kutta solver (which will give you all the positions you need in a single go). The choice between the two is a tradeoff between numerical cost and ease of implementation, and there is no clear recommendation in general.
One thing you should keep in mind is that it's not a terrible thing to be solving differential equations for numerical evaluation. In fact, this is very often done under the hood when you numerically evaluate e.g. Bessel, Airy or Legendre functions. Additionally, it is much more useful as practice for the future to implement a sturdy ODE solver than to embed a Newton's-method solution inside a big algebraic kludge. But again, this comes down to taste.
Additionally, one important reason why the numerical solution is very often the one that's implemented is that it is much more flexible in terms of its growth. For one, it is naturally poised in terms of cartesian coordinates, which you will likely find easier to handle, and it expands more easily to accommodate additional particles or any extra forces you might want to put in in the future. Again, it's up to what you want to use your codebase for in the future. For a planetary system with multiple planets, though, at some point you will want to worry about interactions between the planets, and you may want to explore alternative scenarios with stronger coupling between them.
Finally, I'll note that if all you want is a plot of the orbit, without any time dependence, then you can simply implement equation $(1)$ directly and it will give you all the geometrical information.
