I'm interpreting your question as follows: when you write $H = \pi_1(Y,\ast)$, you mean the inclusion $i:Y\rightarrow X$ induces an injective map on $\pi_1$ with image $H$. (As opposed to $\pi_1(Y)$ being abstractly isomorphic to $H$).
With this interpretation, the answer is no. For example, consider $X = S^1$. It is well known that $\pi_1(X) = \mathbb{Z}$. Now, consider the subgroup $H = 2\mathbb{Z}$. I claim that there is no proper subspace $Y$ for which which has this fundamental group.
We may assume wlog that $Y$ is connected. Then note that any connected proper subset of $S^1$ is homeomorphic to a connected subset of $(0,1)$. These are easy to classify, they are, up to homeomorphism, $(0,1)$, $[0,1]$, and $(0,1]$. None of these has infinite cyclic fundamental group.
Edit Here is an example with $H$ not even abstractly isomorphic to a particular subgroup of $\pi_1(X)$.
Take $X = S^1 \vee S^1$, the wedge sum of 2 $S^1$s. The fundamental group of $X$ is known to be isomorphic to the free group on two generators. It's also know that a free group on two generators contains subgroups isomorphic to the free group on $n$ generators for any finite $n$ (we make even take $n$ to be countable). Further, it's know that free groups on different numbers of generators are never isomorphic.
Let $H$ denote any of these subgroups for $n > 2$. In particular, $H$ is not isomorphic to either $0$, $\mathbb{Z}$, or the free group on two generators. I claim that no subspace $Y$ of $X$ has $H$ as a fundamental group, even up to abstract isomorphism.
As above, we may assume $Y$ is connected. If $Y$ does not contain the wedge point, then it must be contained in a proper portion of one of the two circles, so the above argument shows $\pi_1(Y)$ is trivial. Hence, $Y$ must contain the wedge point. Now, if $Y$ does not contain the whole of the first circle, then $Y$ deformation retracts onto a subspace of the other circle, hence, by the previous argument has $\pi_1 = 0$ or $\mathbb{Z}$. Thus, $Y$ must contain the while first circle. Likewise, $Y$ must contain the whole second circle.
But then this implies $Y = X$, so $Y$s fundamental group is isomorphic to the free group on $2$ generators, so not isomorphic to $H$.
Final (?) Edit Here's one in the finite fundamental group case. Let $X$ be obtained from $S^1$ by attaching a $D^2$ by a degree $4$ map. A simple van Kampen argument shows $\pi_1(X) = \mathbb{Z}/4$. Let $H$ be the unique subgroup of $\pi_1(X)$ isomorphic to $\mathbb{Z}/2$.
I claim that no subspace $Y$ has fundamental group abstractly isomorphic to $H$. If $Y$ misses a point of the interior of the $D^2$, then $Y$ deformation retracts onto a subspace of $S^1$, so by the above argument, has $\pi_1 = 0$ or $\mathbb{Z}$. Hence, we may assume wlog that $Y$ contains all of the interior of $D^2$. Likewise, if $Y$ misses a point of $S^1$, then restricting the van Kampen argument to $Y$ shows that $\pi_1(Y) = 0$, so $Y$ must contain all of $S^1$. This implies $Y = X$, so $\pi_1(Y)\neq H$.
I don't have a complete answer for you, but I can say a bit.
It's a general theorem that if we have a covering space $p:\tilde{X}\rightarrow X$ with $\tilde{X}$ simply-connected and $X$ path-connected and locally path-connected, then the group of deck transformations $G$ is isomorphic to $\pi_1(X)$. (For example, see Hatcher p.71 prop. 1.39.) All the hypotheses hold for this covering $S^3\rightarrow S^3/I^*=P_3$, so this shows that $I^*=\pi_1(P_3)$.
The rest is pure group theory. First look at the icosahedral rotational group, call it $I$. $I$ is a simple group of order 60. The commutator subgroup is always normal, so $[I,I]$ is either trivial or all of $I$. $[I,I]$ is not trivial because $I$ is not abelian, so $I=[I,I]$.
Now the center $Z$ of $I^*$ is a two-element group, and $I^*/Z = I$. Clearly $[I^*,I^*]$ maps onto $[I,I]$. So $I^*/[I^*,I^*]$ is either trivial or has two elements.
The correct answer is that $[I^*,I^*]=I^*$ (wikipedia), but I'm afraid I don't know a proof. Showing that $-1\in I^*$ is a product of commutators would do the trick.
A friend referred me to Weibel's book An Introduction to Homological Algebra. Section 6.9, pp.198-199, "Universal Central Extensions", proves (Example 6.9.1 plus Lemma 6.9.2) that $I^*$ is perfect, i.e., equal to its commutator subgroup. The key fact is that $I^*$ is a universal central extension of $I$. For this fact, Weibel refers us to Suzuki, Group Theory I. I don't have that handy.
However, playing with my dodecahedral paperweight suggests a proof strategy. Let $\tilde{a}$ and $\tilde{b}$ be the two generators of $I^*$ in the presentation $\langle \tilde{a},\tilde{b} | \tilde{a}^5 = \tilde{b}^3 = (\tilde{a}\tilde{b})^2\rangle$. Their images in the icosahedral group are $a$, a rotation of $72^\circ$, and $b$, a rotation of $120^\circ$ (and $ab$ is a rotation of $180^\circ$). In $I$, $a^5=b^3=(ab)^2=1$. Since 1 in $I$ has two preimages in $I^*$, namely $\pm 1$, we must have $\tilde{a}^5 = \tilde{b}^3 = (\tilde{a}\tilde{b})^2 = -1$. (If it equalled 1, then we'd get $I$ instead of $I^*$.)
Now $[a,b]$ is a rotation of $72^\circ$ in $I$ ("proof by paperweight"). So it's conjugate to $a$ in I, and thus $[a,b]^5=1$. That suggests that $[\tilde{a},\tilde{b}]^5$ should equal $-1$ in $I^*$. It should be possible to verify this by direct computation with $2\times 2$ matrices in $SU(2)$.
Hatcher's Example 2.38, "An Acyclic Space", p.142, is also interesting to look at.
Best Answer
There are some conditions that guarantee that the fundamental group of a space will be abelian. For example, if the fundamental group of an H-space is abelian. In these cases, the first homology group will be isomorphic to the fundamental group (if the space is path connected).
Otherwise, if you're only given the data of $H_1(X)$, you cannot compute $\pi_1(X)$ from that. The reason is simple: if you're given the abelianization of a group $G_{ab}$, the group $G$ could be pretty much anything. It could be the product of $G_{ab}$ with a perfect group, or some other extension of $G$...
To give you an example, a knot in $\mathbb{R}^3$ is almost determined by the fundamental group of its complement (as far as I remember, you also need to specify some orientation -- a knot theorist could correct me if I remember incorrectly). But it's also a theorem that the first homology group of this complement is always $\mathbb{Z}$! Even though knot complements are very well-behaved spaces, you still get a lot with the same first homology group.
In fact the noncommutativity of $\pi_1$ can lead to "strange" situations. For example, if the fundamental group is abelian, then trivial homology ($\tilde{H}_*(X) = 0$) implies trivial homotopy ($\pi_*(X) = 0$). But when the fundamental group is not abelian, then it's not true anymore, and there are in fact tons of so-called acyclic spaces whose homology vanish but who are not contractible. Their fundamental group will be perfect because of the Hurewicz isomorphism, but after that (almost) all bets are off. See for example Acyclic spaces by Dror Farjoun. So not even the whole homology of a space is enough to determine the fundamental group if you don't know that it's abelian.
Another example of possible condition is "$X$ is a co-H-space". The fundamental group of $X$ is then free, so the rank of the abelianization is enough to find $\pi_1$ up to isomorphism. I think you can even find the generators by considering lifts of generators of $H_1$.