Realizing the idea of @NickGill we shall confirm the lower bound for solvable groups with five exceptions of the groups $G$ isomorphic to the groups $C_3,C_5,C_3\times C_3, D_{10}$ and $(C_3\times C_3):C_2={\tt SmallGroup}(18,4)$, which have $f_1(G)$ equal to 1,2,3,3, and 4, respectively.

Theorem 1. If a finite solvable group $G$ is not isomorphic to $C_3,C_5,C_3\times C_3,D_{10}$, or $(C_3\times C_3):C_2$, then $f_1(G)\ge\log_2(|G|)$.

The proof is divides into a series of lemmas and claims.

Lemma 1. Let $H$ be a normal subgroup of a finite group $G$ such that the quotient group is cyclic. Then $f_1(G)\ge f_1(H)+a\cdot |H|+ b$ where $a$ is the largest number such that $\frac12a(a+1)|H|<|G/H|$ and $b$ is the largest number such that $\frac12a(a+1)|H|+(a+1)b<|G/H|$.

Proof. Let $g\in G$ be such that $gH$ is a generator of the cyclic group $G/H$. Let $F\subseteq H$ be a product-1-free set of cardinality $|F|=f_1(H)$. Choose any subset $B\subseteq g^{a+1}H$ of cardinality $|B|=b$ and observe that the set $$E=F\cup B\cup\bigcup_{k=1}^a g^kH$$is product-1-free in $G$ and hence $$f_1(G)\ge |E|=f_1(H)+b+a\cdot|H|.$$

The following inequality was proved by @NickGill in his answer.

Lemma 2. For any normal subgroup $H$ of a finite group $G$ we have $$f_1(G)\ge f_1(H)+f_1(G/H).$$

Lemma 3. For any cyclic group $C_n$ of order $n>5$ we have $f_1(C_n)>\log_2(n)$.

Proof. Let $k=f_1(C_n)$. As we already know, $n\le \frac12(k+1)(k+2)$. Assuming that $k=f_1(C_n)\le \log_2(n)$, we conclude that $2^k\le n\le \frac12(k+1)(k+2)$, which implies $k\le 5$.

Lemma 4. For any cyclic group $C_n$ of order $n\notin\{3,5\}$ we have $f_1(C_n)\ge\log_2(n)$.

Proof. Lemma 4 follows from Lemma 3 and known values of $f_1(C_n)$ for $n\le 5$.

Lemma 5. Let $H$ be a non-trivial normal subgroup of a finite group $G$ such that $G/H$ is cyclic. If $f_1(H)\ge \log_2(|H|)$, then $f_1(G)\ge\log_2(|G|)$.

Proof. If $|G/H|\notin\{3,4\}$, then by Lemmas 2 and 4, $$f_1(G)\ge f_1(H)+f_1(G/H)\ge \log_2(|H|)+\log_2(|G/H|)=\log_2(|G|).$$

Now assume that $|G/H|=3$. Choose the largest number $a$ such that $\frac12a(a+1)|H|<|G/H|=3$ and the largest number $b$ such that $\frac12a(a+1)|H|+(a+1)b<|G/H|=3$. If $a\ge 1$, then $a\cdot|H|+b\ge |H|\ge 2$. If $a=0$, then $b=2$ and $a\cdot |H|+b=2$. In both cases we obtain $a\cdot |H|+b\ge 2$. By Lemma 1, $$f_1(G)\ge f_1(H)+a|H|+b\ge f_1(H)+2\ge \log_2(|H|)+\log_2(3)=\log_2(|G|).$$

Finally, assume that $|G/H|=5$. Choose the largest number $a$ such that $\frac12a(a+1)|H|<|G/H|=5$ and the largest number $b$ such that $\frac12a(a+1)|H|+(a+1)b<|G/H|=5$. If $|H|\ge 5$, then $a=0$, $b=4$, and $a\cdot|H|+b=4$. If $|H|=4$, then $a=1$, $b=0$ and $a\cdot|H|+b=4$. If $|H|=3$, then $a=1$, $b=0$ and $a\cdot|H|+b=3$. If $|H|=2$, then $a=1$, $b=1$ and $a\cdot|H|+b=3$. In all cases we have the inequality $a\cdot|H|+b\ge 3$. By Lemma 1, $$f_1(G)\ge f_1(H)+a\cdot|H|+b\ge f_1(H)+3\ge \log_2(|H|)+\log_2(5)=\log_2(|G|).$$

Now we can present

Proof of Theorem 1. Assume that $G$ is a finite solvable group, not isomorphic to $C_3,C_5,C_3\times C_3,D_{10}$ or $(C_3\times C_3):C_2$. If $G$ is trivial, then $f_1(G)=0=\log_2(|G|)$ and we are done. So, we assume that $G$ is not trivial. Being finite and solvable, the group $G$ admits a series of normal subgroups $G_0\subset G_1\subset \dots \subset G_n=G$ such that the group $G_0$ is trivial and for every $k\in\{1,\dots,n\}$ the quotient group $G_k/G_{k-1}$ is cyclic of prime cardinality. If $|G_1|\notin\{3,5\}$, then $f_1(G_1)\ge\log_2(|G_1|)$ by Lemma 4. Applying Lemma 5, we can inductively prove that $f_1(G_k)\ge\log_2(|G_k|)$ for every $k\in\{1,\dots,n\}$.

It remains to consider the case $|G_1|\in\{3,5\}$. Taking into account that $G=G_n$ is not isomorphic to $C_3$ or $C_5$, we conclude that $n>1$. Choose the largest number $a$ such that $\frac12a(a+1)|G_1|<|G_2/G_1|$ and the largest number $b$ such that $\frac12a(a+1)|G_1|+(a+1)b<|G_2/G_1|$. Lemma 1 implies that $f_1(G_2)\ge f_1(G_1)+a|G_1|+b$.

Claim 1. If $G_2$ is not isomorphic to $C_3\times C_3$ or $D_{10}$, then $f_1(G_2)\ge\log_2(|G_2|)$ and $f_1(G)\ge\log_2(|G|)$.

Proof. Assume that the group $G_2$ is not isomorphic to $C_3\times C_3$ or $D_{10}$ but $f_1(G_2)<\log_2(|G_2|)$. Then $f_1(G_1)+a|G_1|+b\le f_1(G_2)<\log_2(|G_2|)$.

The maximality of $a$ implies that $\frac12(a+1)(a+2)|G_1|\ge |G_2/G_1|$ and hence $\frac12(a+1)(a+2)|G_1|^2\ge |G_2|$.

If $|G_1|=3$, then the latter inequality implies $\frac92(a+1)(a+2)\ge |G_2|$. On the other hand, we have the inequality $1+3a\le f_1(G_1)+a|G_1|+b<\log_2(|G_2|)$ which implies $2^{1+3a}<|G_2|\le \frac92(a+1)(a+2)$ and $a\le 1$. Then $|G_2/G_1|\le \frac32(a+1)(a+2)\le 9$. Taking into account that $|G_2/G_1|$ is prime, we conclude that $|G_2/G_1|\in\{2,3,5,7\}$.

If $|G_2/G_1|=7$, then $a=1$, $b=1$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=1+3+1=5>\log_2(21)=\log_2(|G_2|),$$ which contradicts our assumption.

If $|G_2/G_1|=5$, then $a=1$, $b=0$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=1+3+0=4>\log_2(15)=\log_2(|G_2|),$$ which contradicts our assumption.

If $|G_2/G_1|=3$, then the group $G_2$ is has cardinality 9. Since $G_2$ is not isomorphic to $C_3\times C_3$, it is cyclic of order 9. Lemma 4, $f_1(G_2)>\log_2(|G_2|)$, which contradicts our assumption.

If $|G_2/G_1|=2$, then the group $G_2$ is isomorphic to $C_6$ or $S_3$ and hence has $f_1(G_2)=3>\log_2(|G|)$ by the GAP-calculations. But the equality $f_1(G_2)>\log_2(|G|)$ contradicts our assumption.

In all cases we obtain contradictions, showing that $|G_1|\ne 3$.

Then $|G_1|=5$. In this case the inequality $\frac12(a+1)(a+2)|G_1|^2\ge |G_2|$ implies $\frac{25}2(a+1)(a+2)\ge |G_2|$. On the other hand, we have the inequality $2+5a\le f_1(G_1)+a|G_1|+b<\log_2(|G_2|)$ which implies $2^{2+5a}<|G_2|\le \frac{25}2(a+1)(a+2)$ and $a\le 1$. Then $|G_2/G_1|\le \frac52(a+1)(a+2)\le 15$. Taking into account that $|G_2/G_1|$ is prime, we conclude that $|G_2/G_1|\in\{2,3,5,7,11,13\}$.

If $|G_2/G_1|=13$, then $a=1$, $b=3$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=2+5+3=10>\log_2(65)=\log_2(|G_2|),$$ which contradicts our assumption.

If $|G_2/G_1|=11$, then $a=1$, $b=2$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=2+5+2=9>\log_2(55)=\log_2(|G_2|),$$ which contradicts our assumption.

If $|G_2/G_1|=7$, then $a=1$, $b=0$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=2+5+0=7>\log_2(35)=\log_2(|G_2|),$$ which contradicts our assumption.

If $|G_2/G_1|=5$, then $a=0$, $b=4$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=2+0+4=6>\log_2(25)=\log_2(|G_2|),$$ which contradicts our assumption.

If $|G_2/G_1|=3$, then $a=0$, $b=2$ and $$f_1(G_2)\ge f_1(G_1)+a|G_1|+b=2+0+2=4>\log_2(15)=\log_2(|G_2|),$$ which contradicts our assumption.

So, $|G_2/G_1|=2$. Since the group $G_2$ is not isomorphic to the dihedral group $D_{10}$, $G_2$ is cyclic of order 10 and hence $f_1(G_2)>\log_2(|G_2|)$ by Lemma 4.

In all cases we obtained contradictions, showing that the assumption $f_1(G_2)<\log_2(|G_2|)$ was false. Therefore, $f_1(G_2)\ge\log_2(|G_2|)$. Now applying Corollary 2, we can inductively prove that $f_1(G_k)\ge\log_2(|G_k|)$ for all $k\in\{2,\dots,n\}$. This completes the proof of Claim 1.

It remains to consider the cases when the group $G_2$ is isomorphic to $C_3\times C_3$ or $D_{10}$. Since the group $G=G_n$ is not isomorphic to these two groups, we conclude that $n>2$. GAP-calculations show that $f_1(C_3\times C_3)=3=f_1(D_{10})$.

Let $a$ be the largest number such that $\frac12a(a+1)|G_2|<|G_3/G_2|$ and $b$ be the largest number such that $\frac12a(a+1)|G_2|+(a+1)b<|G_3/G_2|$. Then $$\tfrac12a(a+1)|G_2|^2<|G_3|\le \tfrac12(a+1)(a+2)|G_2|^2.$$

Claim 2. If $G_2$ is isomorphic to $D_{10}$, then $f_1(G_3)>\log_2(|G_3|)$ and $f_1(G)>\log_2(G)$.

Proof. To derive a contradiction, assume that $f_1(G_3)\le \log_2(|G_3|)$. By Theorem 1, $$3+10a\le f_1(G_2)+a|G_2|+b\le f_1(G_3)\le \log_2(|G_3|)$$ and hence $2^{3+10a}\le |G_3|$. The maximality of $a$ guarantees that $\frac{100}2(a+1)(a+2)=\frac12(a+1)(a+2)|G_2|^2\ge|G_3|$. Then $2^{3+10a}\le \frac{100}2(a+1)(a+2)$ and hence $a=0$ and $b=|G_3/G_2|-1$. Then $|G_3/G_2|\le \frac12(a+1)(a+2)|G_2|=|G_2|=10$. Taking into account that $|G_3/G_2|$ is prime, we conclude that $|G_2/G_2|\in\{2,3,5,7\}$.

If $|G_3/G_2|\in\{3,5,7\}$, then $$f_1(G_3)\ge f_1(G_2)+a|G_2|+b=3+0+|G_3/G_2|-1>\log_2(|G_3/G_2|\cdot 9)=\log_2(|G_3|),$$which contradicts our assumption. So, $|G_3/G_2|=2$. In this case $|G_3|=20$ and $f_1(G_3)\ge 5>\log_2(|G_3|)$ by the GAP-calculations. Applying Lemma 5, we can inductively prove that $f_1(G_k)>\log_2(G_k)$ for all $k\in\{3,\dots,n\}$.

Claim 3. If $G_2$ is isomorphic to $C_3\times C_3$ but $G_3$ is not isomorphic to $(C_3\times C_3):C_2$, then $f_1(G_3)>\log_2(|G_3|)$ and $f_1(G)>\log_2(G)$.

Proof. To derive a contradiction, assume that $f_1(G_3)\le \log_2(|G_3|)$. By Theorem 1, $$3+9a\le f_1(G_2)+a|G_2|+b\le f_1(G_3)\le \log_2(|G_3|)$$ and hence $2^{3+9a}\le |G_3|$. The maximality of $a$ guarantees that $\frac{81}2(a+1)(a+2)=\frac12(a+1)(a+2)|G_2|^2\ge|G_3|$. Then $2^{3+9a}\le \frac{81}2(a+1)(a+2)$ and hence $a=0$ and $b=|G_3/G_2|-1$. Then $|G_3/G_2|\le \frac12(a+1)(a+2)|G_2|=|G_2|=9$. Taking into account that $|G_3/G_2|$ is prime, we conclude that $|G_2/G_2|\in\{2,3,5,7\}$.

If $|G_3/G_2|\in\{3,5,7\}$, then $$f_1(G_3)\ge f_1(G_2)+a|G_2|+b=3+0+|G_3/G_2|-1>\log_2(|G_3/G_2|\cdot 9)=\log_2(|G_3|),$$which contradicts our assumption. So, $|G_3/G_2|=2$. In this case $|G_3|=18$ and GAP-calclulations show that $f_1(G_3)\ge 5>\log_2(18)=\log_2(|G_3)|$ (as $G_3$ is not isomorphic to $(C_3\times C_3):C_2$. This contradiction shows that $f_1(G_3)>\log_2(|G_3|)$. Applying Lemma 5, we can inductively prove that $f_1(G_k)>\log_2(|G_k|)$ for all $k\in\{3,\dots,n\}$.

Claim 4. If $G_3$ is isomorphic to $(C_3\times C_3):C_2$, then $f_1(G_4)>\log_2(|G_4|)$ and $f_1(G)>\log_2(|G|)$.

Proof. Since $G_3$ is isomorphic to $(C_3\times C_3):C_2$ and $G$ is not isomorphic to this group, $n>3$. To derive a contradiction, assume that $f_1(G_4)\le \log_2(|G_4|)$. GAP-calculations show that $f_1(G_3)=f_1(C_3\times C_3):C_2)=4$. Let $a$ be the largest number such that $\frac12a(a+1)|G_3|<|G_4/G_3|$ and $b$ be the largest number such that $\frac12a(a+1)|G_3|+(a+1)b<|G_4/G_3|$.

By Lemma 1, $$f_1(G_4)\ge f_1(G_3)+a|G_3|+b=4+18a+b$$and hence $$2^{4+18a+b}\le 2^{f_1(G_4)}\le |G_4|.$$ The maximality of $a$ guarantees that $$\frac12(a+1)(a+2)|G_3|^2\ge |G_4|\ge 2^{4+18a}$$and hence $a=0$ and $b=|G_4/G_3|-1$. Then $$2^{4+b}=2^{4+18a+b}\le |G_4|=|G_3|\cdot (|G_4/G_3|)=18(b+1)$$ and hence $b=1$. Then $|G_4/G_3|=b+1=2$ and $|G_4|=36$. GAP-calculations show that $f_1(G_4)\ge 6>\log_2(36)=\log_2(|G_4|)$. But this contradicts our assumption.

Therefore, $f_1(G_4)>\log_2(|G_4|)$. Applying Lemma 5, we can inductively prove that $f_1(G_k)>\log_2(|G_k|)$ for all $k\in\{4,\dots,n\}$.