Yes, your $P(X) \propto \exp(a\cdot x +b\cdot x^2 + c\cdot x^3 +d\cdot x^4)$ maximises the entropy $-\int P(X){\rm log} P(X)dX$ for prescribed first four moments, if the skewness and curtosis lie in a certain range:
M. Rockinger and E. Jondau, Entropy densities with an application to autoregressive conditional skewness and kurtosis (2002).
This simple generalization of the normal distribution holds also if higher moments are prescribed, provided that the highest prescribed moment is even. If, for example, only mean, variance, and skewness, are prescribed, then $d$ would be zero and the distribution fails to normalize. In this case the maximum entropy distribution exists, but it has a more complicated form, as discussed here.
You'll still have to determine the coefficients $a,b,c,d$ from the given first four moments. The cited 2002 paper gives a description of an efficient method.
There may be no solution (typically, if the skewness is too large relative to the kurtosis), meaning that a maximum entropy solution of this simple form does not exist (see Figure 1 of the paper). The combination of a prescribed skewness and kurtosis typically leads to a bimodal distribution (see Figures 2 and 3), which may or may not be desirable for your application.
