When restricted to $0$-$1$ inputs, every $\{+,\times\}$-circuit $F(x_1,\ldots,x_n)$ computes some function $F:\{0,1\}^n\to \mathbb{N}$. To obtain a boolean function, we can just add one fanin-1 threshold gate as the output gate. On input $a\in\{0,1\}^n$, the resulting threshold $\{+,\times\}$-circuit then outputs $1$ if $F(a)\geq t$, and outputs $0$ if $F(a)\leq t-1$; the threshold $t=t_n$ can be any positive integer, which may dependent on $n$ but not on input values. The resulting circuit computes some (monotone) boolean function $F':\{0,1\}^n\to \{0,1\}$.
Question: Can threshold $\{+,\times\}$-circuits be efficiently simulated by $\{\lor,\land\}$-circuits?
Under "efficiently" I mean "with at most a polynomial increase of size." The answer is clear "yes" for threshold $t=1$: just replace $+$ by $\lor$, $\times$ by $\land$, and remove the last threshold gate. That is, $\{\lor,\land\}$-circuits are in fact threshold-$1$ $\{+,\times\}$-circuits. But what about larger thresholds, say, $t=2$?
One can define arithmetic analogues $\#C$ of most boolean circuit classes $C$ by just using $+$ instead of OR, $\times$ instead of AND, and $1-x_i$ instead of $\bar{x}_i$. For example, $\#AC^0$ circuits are $\{+,\times\}$-circuits of constant depth with unbounded fanin $+$ and $\times$ gates, and inputs $x_i$ and $1-x_i$. Agrawal, Allender and Datta have shown that threshold $\#AC^0$ = $TC^0$. (Recall that $AC^0$ itself is a proper subset of $TC^0$; take, say, the Majority function.) In other words, constant-depth threshold circuits can be efficiently simulated by constant-depth $\{+,-,\times\}$-circuits, with just a single threshold gate! Note, however, that my question is about monotone circuits (no Minus "$-$" as gates, and even no $1-x_i$ as inputs). Can one (last) threshold gate be so powerful also then? I don't know this stuff, so any related pointers are welcome.
N.B. There is yet another interesting related result due to Arnold Rosenbloom: $\{+,\times\}$-circuits with just one monotone function $g:\mathbb{N}^2\to\{0,1\}$ as output gate can compute every slice function with $O(n)$ gates. A slice function is a monotone boolean function which, for some fixed $k$, outputs $0$ (resp. $1$) on all inputs with less (resp., more) than $k$ ones. On the other hand, easy counting shows that most slice functions require general $\{\lor,\land,\neg\}$-circuits of exponential size. Thus, one "innocent" additional output gate can make monotone circuits omnipotent! My question asks whether this can also happen when $g:\mathbb{N}\to\{0,1\}$ is a fanin-$1$ threshold gate.
ACTUALIZATION (added 03.11.2014): Emil Jeřábek has shown (via an amazingly simple construction, see his answer below) that the answer is "yes" as long as $t\leq n^c$ for a constant $c$. So, the question remains open only for super-polynomial (in $n$) thresholds.
Usually, in applications, only large thresholds do work: we usually need thresholds of the form $2^{n^{\epsilon}}$ for $\epsilon > 0$. Say, if $F:\{0,1\}^n\to \mathbb{N}$ counts the number of $s$-$t$ paths in graph specified by the $0$-$1$ input, then for $t=m^{m^2}$ with $m\approx n^{1/3}$, the threshold-$t$ version of $F$ solves the existence of a Hamiltonian $s$-$t$ path problem on $m$-vertex graphs (see, e.g. here).
(Added 14.11.2014): Since Emil answered a big portion of my question, and since the case of exponential thresholds is not in sight, I now accept this Emil's (very nice) answer.