As you say, four points (without their associated "$t$" values) are not enough to define a planar cubic Bezier curve.
In fact, 6 points are needed. But, on the other hand, the 6 points have to satisfy certain conditions, or else the required curve does not exist. In your case, you know that the curve exists, because it was used to generate the points in the first place.
The full story is given in this paper: J. Kozak, M. Krajnc, Geometric interpolation by planar cubic polynomial curves, Comput. Aided Geom. Des., 24 (2007), pp. 67-78. You can get a copy of the paper at the authors website.
Another approach (instead of the clever techniques suggested in the paper) is to use brute-force numerical methods. Start with 6 points $X_1, \ldots, X_6$ and 6 parameter values $t_1, \ldots, t_6$. We can assume that $t_1 = 0$ and $t_6 = 1$, but $t_2, t_3, t_4, t_5$ are unknown.We want to find control points $P_1, P_2, P_3, P_4$ for our curve. Obviously we can set $P_1 = X_1$ and $P_4 = X_6$, but $P_2$ and $P_3$ are unknown. We can (numerically) find values of $t_2, t_3, t_4, t_5$ and $P_2$ and $P_3$ that minimize the quantity
$$ \sum_{i=2}^{i=5}\Vert C(t_i) - X_i \Vert^2 $$
where $C(t)$ is the Bezier curve with control points $P_1, P_2, P_3, P_4$. This is a fairly nasty non-linear problem, but a good numerical algorithm will usually converge to the desired solution, given a decent starting point.
Another numerical approach: with the same notation we used above, solve the equations:
$$ C(t_i) = X_i \quad (i = 2,3,4,5)$$
Since the $X_i$ are 2D points, there are 8 equations here. We have 8 unknowns -- $t_2, t_3, t_4, t_5$ and the $x$ and $y$ coordinates of $P_2$ and $P_3$. The equations are non-linear, but a good numerical root-finding package should be able to handle them.
A much better approach is to get the Bezier curves from the CAD package. If the design guys won't give you the curves, you should charge them more money for your services, since they are making your life so much more difficult. Their choice -- curves or cash :-)
The simplest approach is to use $N+1$ points to construct a curve of degree $N$. You have to assign a parameter ($t$) value to each point. So, to construct a cubic curve through four given points $\mathbf{P}_0$, $\mathbf{P}_1$, $\mathbf{P}_2$, $\mathbf{P}_3$, you need four parameter values, $t_0, t_1, t_2, t_3$. Then, as @fang said in his answer, you can construct a set of four linear equations and solve for the four control points of the curve. Two of the equations are trivial, so actually you only have to solve two equations.
The simplest approach is to just set
$$t_0 = 0 \quad , \quad
t_1 = \tfrac13 \quad , \quad
t_2 = \tfrac23 \quad , \quad
t_3 = 1$$
Then the matrix in the system of linear equations is fixed, and you can just invert it once, symbolically. You can get explicit formulae for the control points, as given in this question. But this only works if the given points $\mathbf{P}_0$, $\mathbf{P}_1$, $\mathbf{P}_2$, $\mathbf{P}_3$ are spaced fairly evenly.
To deal with points whose spacing is highly uneven, the usual approach is to use chord-lengths to calculate parameter values. So, you set
$$c_0 = d(\mathbf{P}_0, \mathbf{P}_1) \quad ; \quad
c_1 = d(\mathbf{P}_1, \mathbf{P}_2) \quad ; \quad
c_2 = d(\mathbf{P}_2, \mathbf{P}_3)$$
Then put $c = c_0+c_1+c_2$, and
$$t_0 = 0 \quad , \quad
t_1 = \frac{c_0}{c} \quad , \quad
t_2 = \frac{c_0+c_1}{c} \quad , \quad
t_3 = 1$$
Then, again, provided $t_0 < t_1 < t_2 < t_3$, you can set up a system of linear equations, and solve.
If you're willing to do quite a bit more work, you can actually construct a cubic curve passing through 6 points. Though, in this case, you can't specify the parameter values, of course. For details, see my answer to this question.
Best Answer
some options in R would be
and for your recently added data