Since it is not always clear what $H^1(X;G)$ means for a non-abelian group, Atiyah might have meant this loosely, perhaps using Cech 1-cocycles to construct flat $G$ bundles, with homologous cycles giving isomorphic bundles. The moduli space of flat bundles is homeomorphic (and real analytically isomorphic) to the space of conjugacy classes of $G$ representations via the holonomy. I suspect this is what Ben is hinting at in his answer above.
But another interpretation in terms of homotopy classes of maps to $BG$ is as follows. If you give $G$ the discrete topology, then $BG$ is a pointed space with a fixed choice of isomorphism $\pi_1(BG)\cong G$. So
$Hom(\pi_1(X),G)$ is in bijective correspondence with the pointed homotopy classes of maps $[X,BG]_0$. The free homotopy classes $[X,BG]$ are then in bijective correspondence with the conjugacy classes $Hom(\pi_1(X),G)/conj$, since the action of $G=\pi_1(BG)$ on the pointed homotopy classes corresponds to conjugation, with quotient the free homotopy classes. So if you define $H^1(X;G)=[X,BG]$ (unbased), you get Atiyah's statement.
In the case of Atiyah's book, though, more important that the notation for the space of conjugacy classes of reps is the fact that at a representation $r:\pi_1(\Sigma)\to G$, the (usual, twisted by the adjoint rep) cohomology $H^1(\Sigma; g_r)$ ($g$ the lie algebra of $G$)is the Zariski tangent space to the variety of conjugacy classes of representations at $r$, i.e. to the moduli space of flat $G$ connections on $\Sigma$. The cup product
$H^1(\Sigma; g_r)\times H^1(\Sigma; g_r)\to R$ determines a symplectic form on this variety, and a holomorphic structure on $\Sigma$ induces a complex structure on this variety, which reflects itself in the Zariski tangent space as an almost complex structure $J:H^1(\Sigma; g_r)\to H^1(\Sigma; g_r)$ which coincides with the Hodge $*$ operator on harmonic forms.
Incidentally, all principal $SU(2)$ bundles over a Riemann surface are topologically trivializable.
I figured my comments are getting too long so I should put some of them here, even though I'm using a different topology on $\pi_1$ (often it is the same, but certainly not always) and a possibly different pro-group (which I can't tell as I'm not familiar with toposes).
EDIT: Unless otherwise noted, I'm assuming that $X$ is compact metrizable to be on the safe side.
The relationship goes in 2 steps. First the fundamental pro-group can be compared with the Cech fundamental group endowed with the inverse limit topology. In particular, as noticed by Atiyah and G. Segal, they contain precisely the same information as long as the fundamental pro-group is Mittag-Leffler. For other results in this direction see lemma 3.4 in "Steenrod homotopy". The topological homomorphism between $\pi_1(X)$ (with topology as in my comment) and the topological Cech fundamental pro-group is discussed in theorem 6.1, section 5 and elsewhere in "Steenrod homotopy".
To summarize the relationship very roughly, $\pi_1(X)$ topologized as in my comment retains much of the inverse limit of the fundamental pro-group, discards all of its derived limit, but instead gets something of the derived limit of the second homotopy pro-group (exactly how much is still a subject of ongoing research, see Theorem 6.5 and remark to corollary 8.8 in "Steenrod homotopy"). Still this is not all that it contains (cf. example 5.7 in "Steenrod homotopy").
Added later: Addressing Mike's comment, let me elaborate on the choice of topology on $\pi_1$.
The Mathoverflow thread on the quotient topology on $\pi_1$, and particularly Andrew Stacey's answer there, make it clear that the only reason that this topology is not compatible with multiplication is that the product of two quotient maps need not be a quotient map. But wait, the product of two quotient maps is a quotient map in the category of uniform spaces and uniformly continuous maps! See Isbell's "Uniform spaces" (1964), Exercise III.8(c). In fact, the product of any (possibly infinite) collection of quotient maps is a quotient map in the uniform category. (Quotient maps are defined in any concrete category over the category of sets, as explained e.g. in The Joy of Cats.)
This means that the topology of the quotient uniformity on $\pi_1(X)$ makes it into a topological group. By this topology I mean the following. If $X$ is compact, it carries a unique uniformity, and then the space of continuous (=uniformly continuous) maps $(S^1,pt)\to (X,pt)$ is a uniform space (endowed with the uniformity of uniform conergence; if $X$ is metrizable by a metric $d$, this uniformity is metrizable by the metric $D(f,g)=\sup\limits_{x\in X}\ d(f(x),g(x))$.)
If $X$ is not compact, then we need to fix some uniformity on it and the above works. To what extent the resulting topology on $\pi_1(X)$ depends on the choice of uniformity is a good question.
Added still later: With this in mind, I no longer see a good reason for myself to think at all about the quotient topology on $\pi_1(X)$. As you can see for instance from my comment here, I have good reasons to believe that quotient topology is a sensible notion only when it coincides with the topology of quotient uniformity (i.e. for quotients of compact spaces), and otherwise the topology of quotient uniformity is the way to go. The present situation does not seem to be an exception.
Now does the topology of the quotient uniformity coincide with the "inverse limit" topology, at least when $X$ is compact? I find it easier to go down one dimension, and compare the two similar topologies on $\pi_0(X)$, the set of path-components. The "inverse limit" topology is again induced from the inverse limit topology on the Cech $\pi_0$ (which is the inverse limit of $\pi_0(P_i)$, where $X$ is the inverse limit of the compact polyhedra $P_i$), and alternatively its basic open sets are those collections of path components whose unions are the point-inverses of maps of $X$ to finite sets. And as long as $X$ is compact (or is endowed with a uniformity), we have the topology of the quotient uniformity, viewing $\pi_0(X)$ as the quotient of $X$ by the equivalence relation of being in the same path-component.
Firstly let me say that (often or not) the two topologies do share something. For instance if $X$ is the $p$-adic solenoid then both are anti-discrete, for apparently very different reasons: the inverse limit topology simply because $X$ is connected; and the topology of the quotient uniformity is antidiscrete because every point lies in the closure of the path-component of every other point. If you take a wedge of two solenoids or even a "garland" of solenoids obtained from their disjoint union by identifying pairs of points taken from distincts solenoids, both topologies are still antidiscrete. In the case of the topology of the quotient uniformity, this is so because every two points $x,y$ are the endpoints of a chain $x=p_0,\dots,p_n=y$ such that each $p_{i+1}$ lies in the closure of the path-component of $p_i$.
On the other hand, let us do the pseudo-arc. Its path-components are single points, so $\pi_0(X)=X$. The topology of the quotient uniformity is the original topology of $X$; and the "inverse limit" topology is anti-discrete because $X$ is connected.
I haven't really thought about the two topologies on $\pi_1$ but I see no reason why this should be terribly different from the case of $\pi_0$.
Still more: While the relation of the two topologies on $\pi_1$ might be not so easy to comprehend, the topology of the quotient uniformity on $\pi_1$ relates in a seemingly more comprehensible way directly to the topologized Steenrod $\pi_1$ (whose relation to the fundamental pro-group is clear). The Steenrod $\pi_1(X)$ can be defined as
$\pi_1(holim P_i)$, where the compact metrizable space $X$ is the inverse limit of the polyhedra $P_i$ and $P_0=pt$. (If $P_0\ne pt$ we can shift the indices by one and add a new $P_0$, namely $pt$.) If you think about the construction of $holim$, you will see that $\pi_1(holim P_i)$ is the set of equivalence classes of level-preserving base-ray-preserving maps $S^1\times [0,\infty)\to P_{[0,\infty)}$ under the relation of level-preserving base-ray-preserving homotopy, where $P_{[0,\infty)}$ denotes the mapping telescope of the inverse sequence $\dots\to P_1\to P_0$ (glued out of the mapping cylinders $P_{[i,i+1]}$ of the individual bonding maps). Since $P_0=pt$, "level-preserving" can in fact be replaced by "proper" (see Lemma 2.5 in "Steenrod homotopy"). In other words, the Steenrod $\pi_1(X)$ is the set of equivalence classes of base-ray preserving maps from $D^2$ to the one-point compactification $P_{[0,\infty)}^+$ of the mapping telescope such that the preimage of the point at infinity is precisely the center of the disk (and nothing else).
Now it is not hard to see that the topology on the Steenrod $\pi_1(X)$ (induced from the inverse limit topology on the Cech $\pi_1(X)$ ) is precisely the topology of the quotient uniformity. The uniformity is on a subspace of the space of all continuous (=uniformly continuous) maps from $D^2$ to $P_{[0,\infty)}^+$, and so it is the subspace uniformity (of the uniformity of uniform convergence).
The continuous map from $\pi_1(X)$ with the topology of the quotient uniformity to the Steenrod $\pi_1(X)$ is given by Milnor's lemma (see Lemma 2.1 in Steenrod homotopy): every map $S^1\to X$ extends to a level-preserving map $S^1\times [0,\infty]\to P_{[0,\infty]}$, where $P_{[0,\infty]}$ is Milnor's compactification of the mapping telescope by a copy of $X$, and this extension is well-defined up to homotopy through such extensions. Also close maps have close extensions (from the proof of Milnor's lemma). The quotient $P_{[0,\infty]}/X$ is homeomorphic to $P_{[0,\infty]}^+$, so the close extensions descend to close representatives of elements of the Steenrod $\pi_1$.
Best Answer
No. A continuous homomorphism $S^1\to G$ yields a map $BS^1\to BG$. The space $BS^1$ is homotopy equivalent to $\mathbb CP^\infty$. There is a topological group $G$ such that $BG$ is homotopy equivalent to the sphere $S^2$. A map corresponding to a generator of $\pi_1G=\pi_2BG=H_2S^2$ would give an isomorphism $H^2BG\to H^2BS^1$, but this is incompatible with the cup product.
EDIT: This example is universal in the following sense: A standard way of making a Kan loop group for the suspension of a based simplicial set $K$ is to apply (levelwise) the free group functor from based sets to groups. The realization of this is then the universal example of a topological group $G$ equipped with a continuous map $|K|\to G$. Apply this with $K=S^1$.
EDIT: Yes in the Lie group case. It suffices to consider compact $G$ since a maximal compact subgroup is a deformation retract. Now put a Riemannian structure on $G$ that is left and right invariant, and use that the geodesics are the cosets of the $1$-parameter subgroups, and that in a compact Riemannian manifold every loop is freely homotopic to a closed geodesic.