The answer is no. The smallest example is the presentation of the Klein $4$-group
$$\langle x,y \mid x^2 = y^2 = 1, xy=yx\rangle.$$
In any presentation $\langle S \mid R \rangle$ of this group we have $|R| \ge |S| + 1$.
I don't know whether there is an elementary proof of that. It follows from a more general result that, for a finite presentation $\langle S \mid R \rangle$ of a finite group $G$, we have $|R| \ge |S| + |d(M(G))|$, where $d(M(G))$ is the smallest number of generators of the Schur Multiplier $M(G)$ of $G$.
$M(G)$ is the (unique up to isomorphism) largest group $M$ for which there is a group $E$ with $M \le Z(E) \cap [E,E]$ and $E/M \cong G$. For the Klein 4-group, we have $|M(G)| = 2$, where, for the covering group $E$ we can take either $D_8$ or $Q_8$.
Interestingly, $Q_8$ does have a balanced presentation (i.e. $|S|=|R|$), namely $\langle x,y \mid x^2=y^2, y^{-1}xy=x^3\rangle$.
There is a lot of literature on this topic. Search for balanced presentations, and the deficiency of a presentation.
It's not a convention. It's what the meaning of a presentation is.
When we give a presentation,
$$\langle x_1,x_2,\ldots\mid R_1,R_2,\ldots\rangle$$
(where $R_i$ are the relations satisfied by the generators), we are describing "the most general group that is generated by elements $x_1,x_2,\ldots$ satisfying the relations $R_1$, $R_2,\ldots$". As such, when we write
$$\langle a\mid a^n=e\rangle$$
we are saying "the most general group that is generated by an element $a$ subject to the condition $a^n=e$." The most general such group is the one in which the order is exactly $n$, and not any divisor of $n$.
This idea of "most general group" is captured by von Dyck's Theorem, and by the construction of a group given a presentation.
von Dyck's Theorem. If $G$ is the group given by the presentation
$$\langle x_1,x_2,\ldots\mid R_1,R_2,\ldots\rangle$$
then given any group $H$ and any elements $h_1,h_2,\ldots\in H$ such that, replacing $x_i$ with $h_i$ in the relations $R_1,R_2,\ldots$ results in statements that are true in $H$, there exists a unique group homomorphism $\phi\colon G\to H$ such that $\phi(x_i)=h_i$ for $i=1,2,\ldots.$
That is: any group generated by elements satisfying the relations $R_1,R_2,\ldots$ must be a quotient of $G$.
This is true for the cyclic group of order $n$, but not for the cyclic group of order $k$ for $k\neq n$, $k\mid n$: because, for example, the cyclic group of order $10$ is generated by an element satisfying $x^{20}=1$, but so does the cyclic group of order $20$, and the latter is not a quotient of the former.
Construction. To construct the group given by the presentation given above, first we rewrite all relations so that they are in the form $w_i(x_1,x_2,\ldots)=1$; for example, if the first relation is $x_1x_2=x_2x_1$, then we rewrite it as $x_1x_2x_1^{-1}x_2^{-1}=1$. Then we take the free group $F$ on $x_1,x_2,\ldots$, and let $N$ be the smallest normal subgroup of $F$ that contains all the relations $w_1(x_1,x_2,\ldots), w_2(x_1,x_2,\ldots),\ldots$. Then the group we want is $G=F/N$.
Caveat. Note that just because a presentation has a relations of the form $x^n=e$, that by itself does not mean that the order of $x$ is $n$; it is possible that, when combined with other relations, the order of $x$ will be smaller. For example, as discussed here (and you may want to take a look at that answer anyway), the group
$$G = \langle a,b\mid a^5 = b^4 = 1, aba^{-1}b=1\rangle$$
will actually yield a group in which the order of $b$ is two, not $4$. That's because the other relations, together with the relation $b^4=1$, imply that $b^2=1$ must also hold. So just because you see $x^n=1$ in the presentation, it does not mean that $x$ will definitely have order $n$.
In the case of the cyclic group, $\langle a\mid a^n=1\rangle$, there are no other relations to interact with $a^n=1$, so that one can show that you get the group of order $n$; but this is not a matter of "convention" or "understanding", but of what the presentation requires and means.
Best Answer
Yes. The group is the free group on the generators modulo the normal closure of (smallest normal subgroup generated by) the subgroup generated by the relators.
So in your example, say, $G=F_3/\langle a^2,b^3,c^4,b^{-1}abc^{-1},c^{-1}aba,c^{-1}bca\rangle $.
The group $N$ that we quotient by consists in the smallest normal subgroup containing all the reduced words that can be formed out of the relators.
Note that the same group can have multiple presentations. The group can turn out to be trivial or free, for instance.