If Gerhard Paseman is right that $t(7!)=8$ then $(k,a)=(7,13)$ is already an example of $t(ak!)<t(k!)$, because $$ 13 \cdot 7! = 65520 = 2^{16} - 2^4 $$ can be reached in six steps from $1$: $$ 1 + 1 = 2, \quad 2 \cdot 2 = 4, \quad 4 \cdot 4 = 16, \quad 16 \cdot 16 = 256, \quad 256 \cdot 256 = 65536, $$ and finally $65536 - 16 = 65520$. This also makes $(10,1183)$ a candidate if $10!$ can't be reached as quickly as the seven steps to $65520^2$. [Added later: Michael Stoll now confirms that $t(7!)=8$ $-$ and also reports that it it takes nine steps to reach $10!$, and and eight for $8!$ and $9!$, so this this seven-step route to $65520^2$ gives gives $t(ak!)<t(k!)$ also for $k=8,9,10$. Later yet, Stoll links to OEIS A217032 which gives the values of $t(k!)-1$ for all $k \leq 19$; this lets us give some more examples, such as $16! \, | \, (2^{64}-2^4)^4$ with $t(16!) = 13$ and $t((2^{64}-2^4)^4) \leq 11$.]
But the asymptotic question is much harder because we don't have good lower bounds on $t(k!)$. It's certainly smaller than Gerhard Paseman gives an upper bound of $2k$ or maybe $(1+o(1))k$, which is within a constant factor of what can be done for any number of this size: if $N < k^k$ we can use $k$ additions to get $1,2,3,\ldots,k$, forthen $k-1$ multiplications to reach $k^2, k^3, \ldots, k^k$, and then $k+k$ more multiplications and additions to reach $N$ via its base-$k$ expansion. For $t(k!)$ we can reduce this by a factor $\gg \log k$ for large $k$, because $k!$ is a product of powers powers of the $\pi(k)$ primes $p\leq k$. We can constructreach all those primes in in $\pi(k) + O(k^{1/2})$ additions additions: let $r = \lfloor \sqrt{k} \rfloor$, then then add to get $1,2,3,\ldots,r,2r,3r,\ldots, r^2$, and each prime is a sum of one or two of these. [The Masked Avenger notes that it's a bit faster to first get all the numbers that occur as gaps $p_n - p_{n-1}$, and then get from each prime to the next.] Now use fewer than $\pi(k)$ multiplications to make the products $$ P_i := \prod_{ip \leq k < (i+1) p} p $$ for $i=1,2,3,\ldots,r$, and then $k! = P_1^{\phantom 1} P_2^2 P_3^3 \cdots P_r^r$ times $\pi(r)$ prime powers.
If that approach were optimal then we could get $t(ak!) < t(k!)$ by replacing replacing $\prod_{i=1}^r P_i^i$ by $\left(\prod_{i=1}^r P_i\right)^{2^\rho}$ once $2^\rho \geq r$. But it seems that in fact $t(k) \ll k^{1/2 + o(1)}$ because one can write $m^2! = \prod_{j=0}^{m-1} Q(jm)$ where $Q(X) = (X+1) (X+2) (X+3) \cdots (X+m)$ and use FFT-like tricks to evaluate a degree-$m$ polynomial at $m$ points in only $m \log^A m$ operations. (See for instance this page. I learned of this surprising fact some year back from Henry Cohn. Caveat: some more work might be needed to fit this method into the $\{+,-,\times\}$ model without losing a factor worse than $k^\epsilon$. Added later: Turbo notes that this paper by Q.Cheng (of which more in the next paragraph) cites a paper by Strassen from the 1970's that gives a route to $k!$ in $O(k^{1/2} \log^2 k)$ $\{+,-,\times\}$ steps.)
This might be asymptotically optimal, but proving $t(k!) \gg k^{1/2}$ seems(or even $t(k) \gg k^\theta$ for some $\theta>0$) seems to be beyond reach. Meanwhile Meanwhile, Lenstra's ECM (elliptic curve method) for for factoring suggests suggests that $t(ak!)$ can be as small as $\exp O(\sqrt{\log k\,})$. Turbo gave this link to a 2004 paper by Qi Cheng that spells out this connection; that's perhaps surprisingly late, since Lenstra's ECM paper dates back to 1987 $-$ I found some e-mails from 1996 where I noted that ECM suggests that some multiple of $k!$ can be computed in a number of operations subexponential in $\log k$, and I wouldn't be surprised if Lenstra himself noticed this some years earlier. Note that likewise the counterexamples involving $2^{16}-2^4$ etc. that I started with are in effect using Pollard's $p-1$ factorization method.
I tried to estimate how well this would work for $k = 10^7$
compared with the prime-factorization technique.
There are $664579$ primes $p < 10^7$, so any route to $k!$ via
prime factorization would have to take at least $2 \cdot 664579$ steps
(one to reach each $p$ and one to multiply by it).
For the ECM approach to some multiple of $k!$, I tried the following experiment.
For each of the first $96$ isogeny classes of rank-$1$ elliptic tables in
Cremona's table (these being the first two columns of Table 2 on page 235 of
his Algorithms for Modular Elliptic Curves, covering conductors $N \leq 220$),
choose a curve $E_i$ and a generator $P_i$
($i \leq 96$). Then, for each prime $p \leq 10^7$, factor the order of
each $P_i \bmod p$ using the built-in gp command ellorder,
factor it as $\prod_j l_j^{e_j}$, and note the minimal value $m(p)$ of
$\max_j l_j^{e_j}$ over the $96$ choices of $(E_i,P_i)$;
thus for each $p$ there's some $i$ for which the order of $P_i \bmod p$
is a product of prime powers $\leq m(p)$.
This took a few hours.
The largest $m(p)$ observed is $379$, for $p = 6978421$.
But about 90% of these primes have $m(p) \leq 83$.
(There are $67608$ primes $p\leq 10^7$ with $m(p)>83$;
also $47193$ with $m(p)>100$ and $22518$ with $m(p)>113$.)
There are $23$ primes $l \leq 83$; let $M = 2^6 3^4 5^2 7^2 \prod_{l=11}^{83} l$,
and compute for each $i$ a nonzero multiple $D_i$ of the denominator of $M P_i$,
which requires about $150$ group-law additions in $E_i$ because $M \approx 2^{122.6}$.
Then form $\prod_{i=1}^{96} D_i$, multiply by the product of the $67608$
missing primes, and square the whole thing $23$ times to get a multiple
of $10^7!$. I don't know how many arithmetic operations it takes these days
to do an elliptic-curve addition or subtraction, but it must be under $40$,
and this already puts us under half our estimate for computing $10^7!$ itself
via prime factorization ($96 \cdot 150 \cdot 40 = 600000$). There's probably a significant additional factor to be saved by
using more curves $E_i$, balancing the $m(p)$ cutoff, and choosing
$E_i$ that allow for faster group operations such as Edwards curves.