These are indeed equivalent!
The proof of your 3.$\Rightarrow$1. requires more work than the other implications though. I recommend you have a look at Theorem 2.4.1 in
Kreuzer & Robbiano's Computational Commutative Algebra 1, where they list over ten equivalent conditions on a set $\{g_1,\dots,g_s\}$ to be a Gröbner bases for the ideal $I$ that they generate.
They are of four flavours: Ideal membership certificates, initial ideal generation, confluency of rewrite rules, and lifting of syzgies:
$A)$ (close to your 2.) For every $0 \neq m \in I$ there exists $f_1,\dots,f_s\in P$ with $m = \sum_i f_ig_i$ and $\mathrm{LT}(m) \succ\mathrm{LT}(f_ig_i)$ for all $f_ig_i \neq 0$.
$B)$ (your 1.) The set $\{\mathrm{LT}(g_1),\dots,\mathrm{LT}(g_s)\}$ generates the initial ideal $\mathrm{LT}(I)$.
$D)$ Every syzygy in $\operatorname{Syz}(\mathrm{LT}(g_1),\dots,\mathrm{LT}(g_1))$ lifts to a syzygy in $\operatorname{Syz}(g_1,\dots,g_1)$ (we don't need this in the sequel so I won't elaborate what this means precisely).
Before stating the conditions $C_i)$, which are close to your 3., introduce the following notation/terminology:
- Write $m_1 \overset{g_i}{\longrightarrow} m_2$ if there is a term $c x^\alpha$ and an index $i$ such that $m_2 = m_1 - cx^\alpha g_i$ and $x^\alpha \mathrm{LT}(g_i) \notin \mathrm{supp}(m_2)$. Passing from $m_1$ to $m_2$ in this way is a reduction step (via $g_i$).
- Let $\overset{G}{\longrightarrow}$ be the rewrite relation on $P$ that is the "joint" transitive hull of ${\overset{g_1}{\longrightarrow}},\dots,{\overset{g_s}{\longrightarrow}}$.
- $m$ is irreducible (w.r.t. $G$) if $m \overset{G}{\longrightarrow} m'$ implies $m'=m$ (not to be confused with irreducible in the sense of factoring, unfortunate terminology).
$C_1)$ For any $m \in P$ we have $m \overset{G}{\longrightarrow} 0$ if
and only if $m \in M$.
$C_2)$ If $m \in I$ is irreducible then $m=0$.
$C_3)$ (your 3.) For any $m_1 \in P$ there is a unique irreducible $m_2 \in P$
such that $m_1 \overset{G}{\longrightarrow} m_2$.
$C_4)$ (Confluency) If $m_1 \overset{G}{\longrightarrow} m_2$ and $m_1 \overset{G}{\longrightarrow} m_3$, then there exists a $m_4$ with $m_2\overset{G}{\longrightarrow}m_4$ and $m_3\overset{G}{\longrightarrow}m_4$ (picture this as a diamond-shaped graph!).
Let me sketch the essential ideas on the equivalence of the $C_i)$ (Proposition 2.2.5) and the connection to $A),B)$ (Proposition 2.2.8). I invite you to try yourself (hence the spoiler tag), or check out the excellent book.
$C_1)$$\Rightarrow$$C_2)$:
If $m \in I$ and then $m \overset{G}{\longrightarrow} 0$ by $C_1)$, so if $m$ is also irreducible, then $m=0$.
$C_2)$$\Rightarrow$$C_3)$:
The existence of such a $m_2$ follows by reducing repeatedly which must eventually terminate in an irreducible
element as $\succ$ is a well-order. If $m_2'$ is another such element,
then from $m_1 \overset{G}{\longrightarrow} m_2$ and $m_1 \overset{G}{\longrightarrow} m_2'$ you can see that $m_2-m_2' \in I$.
Also, $m_2-m_2'$ is also irreducible (if it was not, then one of $m_2$
or $m_2'$ can't be irreducible either), so by $C_2)$ we get
$m_2=m_2'$.
$C_3)$$\Rightarrow$$C_4)$:
For $m \in P$ let $m^*$ be the unique element from $C_3)$. If $m_1 \overset{G}{\longrightarrow} m_2$ and
$m_1 \overset{G}{\longrightarrow} m_3$, then by uniqueness it follows
$m_2^* = m_1^* = m_3^*$. So $m_4 := m_1^*$ has the desired property.
$C_4)$$\Rightarrow$$C_1)$ (arguably the trickest part of the cycle of implications):
It's clear that $m \in I$ if $m \overset{G}{\longrightarrow} 0$, so we only need to show the converse.
From any ideal membership certificate $m = \sum_i f_ig_i$ one can
deduce that $m \overset{G}{\longleftrightarrow} 0$, where $m\overset{G}{\longleftrightarrow}m'$ means that there is a sequence of
polynomials $(m=m_1,m_2,\dots,m_l=m')$ such that $m_i\overset{G}{\longrightarrow} m_{i+1}$ or $m_{i+1}\overset{G}{\longrightarrow} m_{i}$ ("you can reduce forwards and
backwards"). Let $\ell$ be the largest index such that $m_\ell\overset{G}{\longleftarrow} m_{\ell+1}$ is a backwards step, then
$m_{\ell+1} \overset{G}{\longrightarrow} 0$ and $m_{\ell+1}\overset{G}{\longrightarrow} m_\ell$. By $C_4)$ and the fact that $0$
is irreducible, it follows that $m_\ell \overset{G}{\longrightarrow}0$. We win by induction on the number of backwards steps.
You said you're already familiar with $B)$$\Rightarrow$$C_3)$, so let's finish with $C_2)$$\Rightarrow$$B)$:
Let $0 \neq \mathrm{LT}(m) \in \mathrm{LT}(I)$. By $C_2)$ $m$ is not irreducible, so $m \overset{g_i}{\longrightarrow} m'$ with $m' \neq m$. If the reduction step canceled the leading term, then $\mathrm{LT}(g_i) \mid \mathrm{LT}(m)$ and we win. If not, replace $m$ by $m' \neq 0$ and repeat. As before, this procedure must terminate by the well-ordering of $\prec$.
