Your definitions are in fact those for proper, orthochronous Lorentz transformation, not for general Lorentz transformations, that's why you're having trouble telling the difference! (If it makes you feel any better, yesterday a collegue and I were trying to debug his test setup and two hours of complex testing passed before we two geniusses realised we hadn't switched the power to a key bit of kit on!)
A general Lorentz transformation is defined by criterion 1) alone - it is simply any linear transformation that preserves the quadratic form $t^2 - x^2 - y^2 - z^2$.
The proper, orthochronous transformations are those that belong to the identity connected component $SO^+(1,\,3)$ of the full Lorentz group $O(1,\,3)$. That is, the proper, orthochronous transformations are those that can be reached from the $4\times4$ identity matrix by following a continuous path through the Lorentz group. Equivalently, they are the matrices that are on paths through the Lorentz group defined by the differential equation:
$$\begin{array}{lcl}\mathrm{d}_s L &=& (a_x(s)\, J_x + a_2(s)\, J_y+a_z(s)\, J_z + b_x(s)\, K_x + b_y(s)\, K_y+b_z(s)\, K_z)\,L\\L(0) &=& \mathrm{id}\end{array}\tag{1}$$
where $\mathrm{id}$ is the $4\times 4$ identity, $a_j(s),\,b(s)$ are continuous functions of the parameter $s$ and the $J_j,\,K_J$ are six matrices $4\times 4$ that span the Lie algebra of the Lorentz group, i.e. the real vector space of all possible "tangents to the identity", i.e. all possible values of $\mathrm{d}_s L|_{s=0}$. One possible set is:
$$\begin{array}{lcllcllcl}J_x&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&1&0\end{array}\right)&J_y&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&0&1\\0&0&0&0\\0&-1&0&0\end{array}\right)&J_z&=&\left(\begin{array}{cccc}0&0&0&0\\0&0&-1&0\\0&1&0&0\\0&0&0&0\end{array}\right)\\K_x&=&\left(\begin{array}{cccc}0&10&0&0\\1&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)&K_y&=&\left(\begin{array}{cccc}0&0&1&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{array}\right)&K_z&=&\left(\begin{array}{cccc}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{array}\right)\end{array}\tag{2}$$
(See how the $J_j$ are skew-Hermitian, thus have pure imaginary eigenvalues, so that $\exp(a_j\, J_j)$ has stuff like $\sin,\,\cos$ of an angle and is a rotation matrix, whereas the $K_j$ are Hermitian, with purely real eigenvalues, so that $\exp(b_j\, K_j)$ has stuff like $\sinh,\,\cosh$ of a rapidity and is a pure boost matrix).
An intuitive description: imagine you are sitting at the console of your spaceship's "hyperdrive": it has two track balls each with their own levers marked "spin" and "boost" and a set of accelerometers - linear and rotational. Your spaceship is initially moving inertially. You roll the trackballs around to set the axis of rotation and direction of boost respectively. When you pull on the levers, the spin lever accelerates the angular speed about the rotation axis, the boost lever accelerates the linear velocity in the boost direction. Otherwise put, the "rotate" trackball and its lever set the superposition weights $a_j(s)$ of the $J_j$ in (1) when we use the definitions in (2) and the "boost" trackball and its lever set the weights $b_j$ of the $K_k$. You go through a control sequence, ending so that your accelerometers read nought, so that now a set of $x,\,y,\,z$ axes attached to your spaceship is moving inertially relative to the beginning frame. The proper, orthochronous transformations are precisely every transformation between the beginning frame and an inertial frame that you can reach with your controls.
However, there are other transformations possible that preserve the quadratic form $t^2 - x^2 - y^2 - z^2$ that don't fulfill your criteria 2. and 3. but they follow only a "simple" pattern that makes them "not much different" from the identity connected component. A discrete subgroup of the full Lorentz group is $\{\mathrm{id},\,P,\,T,\,P\,T\}$ with
$$P=\text{"parity flipper"} = \mathrm{diag}[1,\,-1,\,-1,\,-1];\\T=\text{''time flipper''} = \mathrm{diag}[-1,\,1,\,1,\,1]$$
With the exception of $\mathrm{id}$, none of these can be reached from the identity by paths fulfilling (1). They belong to different connected components from the identity component $SO^+(1,\,3)$. Indeed, the identity connected component is a normal subgroup of the full Lorentz group $SO(1,\,3)$ and the quotient $O(1,\,3) / SO^+(1,\,3)$ is the little group $\{\mathrm{id},\,P,\,T,\,P\,T\}$. So any full Lorentz transformation can be represented as a proper orthochronous transformation followed by one of $P,\,T$ or $P\,T$. There are four separate connected components to the full Lorentz group. (an aside: $\{\mathrm{id},\,P,\,T,\,P\,T\}$ is the Klein "fourgroup": the only possible group of four elements aside from $\mathbb{Z}_4$).
To sniff out a non-proper or non-orthochronous transformation, you do one of two things:
Compute the matrix's determinant. If it is -1, then you know it has to include one of $P$ or $T$, so it's not proper or not orthochronous. You can further differentiate the $P$ and $T$ cosets by looking at the $L_0^0$ component of the transformation: the $T$ coset has $L_0^0<0$, since such a transformation swaps the roles of the "future" and "past" (actually reflects Minkowsky vector space in the $t=0$ plane).
If the determinant is $+1$, then it may belong to the $P\,T$ coset of $O(1,\,3)$. As in point 1, the $T$ coset and the $P\,T$ coset can be recognised as transformations with $L_0^0<0$
Unfortunately you can't just define the Poincare group to be that, because in the standard treatment it is defined a little bit differently. What you defined is actually the Lorentz group. The Poincare group contains translations as well.
The Lorentz group $O(1,3)$ is the group of all $\Lambda \in GL(4, \mathbb{R})$ such that
$$\Lambda^T \eta \Lambda = \eta,$$
with $\eta = \operatorname{diag}(-1,1,1,1)$. This can be seen as the group of all "changes of orthonormal frames in spacetime".
Remember that one orthonormal reference frame is a set of vectors $\{e_\mu\}$ such that $g(e_\mu,e_\nu)=\eta_{\mu\nu}$. In that sense, given two such frames, the change of frame that takes components in one of them to components in the other is given by these elements.
With this, the proper Lorentz group is $SO(1,3)$ which basically means that you pick all elements of $O(1,3)$ with determinant $+1$. The orthocronous part just means that if $\Lambda \in O(1,3)$ has $\Lambda^0_0 > 0$ then it preserves the sense of time-like vectors.
Some authors seems to include "by default" the orthocronous requirement in the group $SO(1,3)$ (see for example Analysis, Manifolds and Physics by Choquet-Bruhat, vol. 1, page 290). Others leave it separately, so that you end up with a group $SO^+(1,3)$, but this is a question of convention.
Now the Poincare group $P(1,3)$ (which I don't know any standard notation for) is the group of all Lorentz transformations together with all spacetime translations. In other words, we have:
$$P(1,3)= \{(a,\Lambda) : a\in \mathbb{R}^4, \Lambda \in SO(1,3)\}$$
together with the multiplication defined by
$$(a_1,\Lambda_1)\cdot (a_2,\Lambda_2)=(a_1+\Lambda_1 a_2, \Lambda_1\Lambda_2)$$
Think like that: while elements of $SO(1,3)$ relates orthonormal frames with coincident origins, elements of $P(1,3)$ also allows for the shift of origin as well.
The action of $SO(1,3)$ in the Minkowski vector space $\mathbb{R}^{1,3}$ (not to be confused with flat spacetime - this is actually the "model" for spacetime's tangent spaces, which just happens to be possible to identify with spacetime itself in the flat case) is given by usual matrix multiplication, i.e., given $ \Lambda \in SO(1,3)$ you have:
$$\Lambda \cdot v = \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
On the other hand, the action of $P(1,3)$ on the Minkowski vector space $\mathbb{R}^{1,3}$ is characterized by the fact that given $(a,\Lambda)\in P(1,3)$ you have:
$$(a,\Lambda)\cdot v = a + \Lambda v, \quad \forall v\in \mathbb{R}^{1,3}.$$
I don't know if it helps, but people like to compare this to the case in $\mathbb{R}^3$ where you have the rotation group $SO(3)$ and the Euclidean group $E(3)$ comprising rotations in $SO(3)$ with translations in $\mathbb{R}^3$ thus forming the group of rigid motions. This could be seen as the analogous construction in spacetime.
EDIT: Regarding the semi-direct product construction mentioned in comments, recall that given groups $N,H$ with $\varphi : H\to \operatorname{Aut}(N)$ a homomorphism into the group of automorphisms of $N$, we can build the semi-direct product as the set $N\times H$ with the product:
$$(a,b)\cdot (c,d)=(a\varphi(b)(c), bd)$$
the resulting group is denoted $N\rtimes H$. In the particular case it is clear that we have this construction with $N = \mathbb{R}^{1,3}$, $H = SO(1,3)$ and $\varphi : SO(1,3)\to \operatorname{Aut}(\mathbb{R}^{1,3})$ given by $\varphi(\Lambda)(v) = \Lambda v$. Thus
$$P(1,3)=\mathbb{R}^{1,3}\rtimes SO(1,3)$$
Best Answer
The proper orthochronous Lorentz group is, as the name suggest, a group. It may be defined, as you said, as the set of transformations leaving the metric invariant, with unit determinant and positive time component. A usual $\mathbb R^3$ rotation, without involving time, also preserves $ds^2=d\vec r^2-(c\,dt)^2$. Other known type of transformation that doesn't change $ds^2$ is \begin{equation}t\,\to\,\gamma(t-v\,x),\qquad x\,\to\, \gamma(x-v\,t),\tag{1}\end{equation} with $\gamma\equiv1/\sqrt{1-v^2}$ (all of this with $c=1$).
I'd say that "Lorentz transformation" should be the name for any transformation inside the Lorentz group, which could be any of the previous ones, or a combination of them. "Boost", "pure boost" or "pure Lorentz transformation" are names usually given only to the type of transformation in (1). Notice that this would be a boost along the $x$-axis, but you could have boost defined in any arbitrary direction of your space.