The paper An update on Hurwitz groups by Marston Conder seems to suggest that the Chevalley group $G(2,5)$ of order $5859000000$ is a quotient of $G := \langle a, b \  \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$. Can we check this, and if it is true, will it help with finding more quotients of the group?
Yes, $G(2,5)$ is a quotient of your group $G$.  We can find generators $a$ and $b$ of $G(2,5)$ satisfying the relations in a few seconds with GAP (so in particular there is no need to buy Magma for this!):
gap> G25 := SimpleGroup("G2(5)");
G2(5)
gap> cl2 := G25.1^G25;; # conjugacy class of involutions
gap> cl3 := G25.2^G25;; # one of the 2 conjugacy classes of elements of order 3
gap> repeat # find the desired generators
> a := Random(cl2); b := Random(cl3);
> until Order(a*b) = 7 and Order(Comm(a,b)) = 10 and G25 = Group(a,b);
gap> Order(a);
2
gap> Order(b);
3
gap> Order(a*b);
7
gap> Order(Comm(a,b));
10

1$\begingroup$ You could also do without charge with the Magma calculator! $\endgroup$ – Derek Holt Nov 21 '13 at 11:58
Yes, it's not hard to check! Just choosing random elements of $G_2(5)$ of order 2 and 3 in Magma, I very quickly found two matrices that satisfy the relations and generate $G_2(q)$:
[4 0 0 0 0 0 0]
[0 4 0 0 0 0 0]
[0 0 1 0 0 0 0]
[0 0 0 1 0 0 0]
[0 0 0 0 1 0 0]
[0 0 0 0 0 4 0]
[0 0 0 0 0 0 4]
[0 0 4 3 1 0 0]
[3 1 4 3 1 3 1]
[4 3 3 1 2 1 2]
[1 2 0 4 1 0 0]
[3 3 0 4 1 0 2]
[1 3 3 4 4 0 4]
[1 4 3 1 1 1 2]
A good reference for these types of calculations is W. Plesken and D. Robertz, Representations, commutative algebra, and Hurwitz groups, J. Algebra 300 (2006), 223247. They find all characteristic zero representations of the Hurwitz group up to degree 7.
They say there that there are two 7dimensional representations of $\langle a,b \mid a^2=b^3=(ab)^7=[a,b]^{10}=1 \rangle$ defined over quadratic extensions of the number field ${\mathbb Q}(\sqrt{5})$. That means that there are finite 7dimensional representations over fields of order $p$ or $p^2$ or $p^4$ for almost all primes $p$. I don't know exactly what the images would be  probably linear or unitary groups for most primes  I can try and find out.
But in any case, there are definitely infinitely many finite simple images of the group.
Added: I have managed to compute a presentation of $G_2(5)$ by introducing a third generator $z$ and two long relations involving $y$ and $z$. It won't fit your idea of a simple presentation, but that's a limitation of computer calculations! The subgroup $\langle y,z \rangle$ has order 744000 and index 7875, so I got the presentation by first getting one for the subgroup and then doing coset enumeration over the subgroup.
< x,y,z  x^2, y^3, (x*y)^7, (x,y)^10, z=x*y*x*y^1*x,
y^1*z*y*z*y^1*z*y*z*y*z*y*z*y^1*z*y^1*z*y*z*y^1*z*y^1*z*y^1*z*y*z*
y*z*y*z*y^1*z*y^1*z*y^1*z*y*z*y^1*z*y*z*y^1*z* y*z*y^1*z,
y^1*z*y*z*y^1*z*y*z*y^1*z*y*z*y^1*z*y*z*y^1*z*y*z*y^1*z*y^1*z*y^1*
z*y^1*z*y*z*y*z*y^1*z*y*z*y^1*z*y*z*y^1*z*y*z*y^1*z*y^1*z*y*z*y*z*
y*z*y*z >;
Added: I have calculated explicitly the two 7dimensional representations over number fields now, and tried reducing modulo some finite prime powers. The finite images all seem to be $G_2(p^e)$ for $e \le 4$. As well as $G_2(5)$,I've found $G_2(3^4)$, $G_2(7^2)$, $G_2(11^2)$, $G_2(13^4)$.

$\begingroup$ Is there a simple presentation of the group as a quotient of G? $\endgroup$ – Thomas Nov 21 '13 at 10:57

$\begingroup$ I don't know. The problem is that the group is too big to easily test computationally whether a candidate presentation is correct. $\endgroup$ – Derek Holt Nov 21 '13 at 11:28

$\begingroup$ So, you're saying that G(2,49), G(2,81), G(2,121), and G(2,28561) are all quotients of G? $\endgroup$ – Thomas Nov 22 '13 at 1:48

$\begingroup$ Yes, that's right. It looks as though $G(2,p^k)$ is a quotient of $G$ $k \le 4$ for all primes, possibly with one or two of exceptions (I don't think it works for $p=2$). $\endgroup$ – Derek Holt Nov 22 '13 at 8:13