Algorithm to test whether a language is regular
The answer is no. Deciding whether a given context-free grammar generates a regular language is an undecidable problem. 

Update. I gave this negative answer to the general question


  Given a language specified in algebraic form, test whether the
  language is regular or not


since context-free languages are solutions of algebraic equations in languages: see Chapter II, Theorems 1.4 and 1.5 in the book of J. Berstel Transductions and Context-Free Languages.

However, the same question is decidable for deterministic context-free languages, a nontrivial result due to Stearns [1] and improved by Valiant [2]:
[1] R. E. Stearns, A Regularity Test for Pushdown Machines, Information and Control  11 323-340 (1967). DOI:10.1016/S0019-9958(67)90591-8.
[2] L. G. Valiant. Regularity and related problems for deterministic pushdown automata J. ACM 22 (1975), pp. 1–10.

There is another positive result, closer to the specifications given in the second part of the question. Recall that the semilinear subsets of $\mathbb{N}^k$ are exactly the sets definable in Presburger arithmetic. There are also the rational subsets of $\mathbb{N}^k$. In particular, a subset of $\mathbb{N}^k$ defined by linear inequations is rational. Now, given a rational subset $R$ of $\mathbb{N}^k$, it is decidable whether the language
$$
  L(R) = \{ u_1^{n_1} \dotsm  u_k^{n_k} \mid (n_1, ...,n_k) \in R \}
$$
is regular. Indeed, it is known [Ginsburg-Spanier] that $L(R)$ is regular if and only if $R$ is a recognizable subset of $\mathbb{N}^k$ and it is decidable [Ginsburg-Spanier] whether a given rational subset of $\mathbb{N}^k$ is recognizable. 

S. Ginsburg and E. H. Spanier., Semigroups, Presburger formulas, and languages,  Pacific J. Math. 16 (1966), 285-296.

S. Ginsburg and E. H. Spanier. Bounded regular sets, Proc. of the American Math. Soc. 17, 1043–1049 (1966).

This does not solve the second part of the question, which might be undecidable because of the word variables, but it gives a reasonable fragment to start with.